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What is organic chemistry? 



1 



Organic chemistry and you 



You are already a highly skilled organic chemist. As you read these words, your eyes are using an 
organic compound (retinal) to convert visible light into nerve impulses. When you picked up this 
book, your muscles were doing chemical reactions on sugars to give you the energy you needed. As 
you understand, gaps between your brain cells are being bridged by simple organic molecules (neuro- 
transmitter amines) so that nerve impulses can be passed around your brain. And you did all that 
without consciously thinking about it. You do not yet understand these processes in your mind as 
well as you can carry them out in your brain and body. You are not alone there. No organic chemist, 
however brilliant, understands the detailed chemical working of the human mind or body very well. 

We, the authors, include ourselves in this generalization, but we are going to show you in this 
book what enormous strides have been taken in the understanding of organic chemistry since the 
science came into being in the early years of the nineteenth century. Organic chemistry began as a 
tentative attempt to understand the chemistry of life. It has grown into the confident basis of vast 
multinational industries that feed, clothe, and cure millions of people without their even being 
aware of the role of chemistry in their lives. Chemists cooperate with physicists and mathemati- 
cians to understand how molecules behave and with biologists to understand how molecules 
determine life processes. The development of these ideas is already a revelation at the beginning of 
the twenty-first century, but is far from complete. We aim not to give you the measurements of the 
skeleton of a dead science but to equip you to understand the conflicting demands of an 
adolescent one. 

Like all sciences, chemistry has a unique place in our pattern of understanding of the universe. It 
is the science of molecules. But organic chemistry is something more. It literally creates itself as it 
grows. Of course we need to study the molecules of nature both because they are interesting in their 
own right and because their functions are important to our lives. Organic chemistry often studies life 
by making new molecules that give information not available from the molecules actually present in 
living things. 

This creation of new molecules has given us new materials such as plastics, new dyes to colour our 
clothes, new perfumes to wear, new drugs to cure diseases. Some people think that these activities are 
unnatural and their products dangerous or unwholesome. But these new molecules are built by 
humans from other molecules found on earth using the skills inherent in our natural brains. Birds 
build nests; man makes houses. Which is unnatural? To the organic chemist this is a meaningless dis- 
tinction. There are toxic compounds and nutritious ones, stable compounds and reactive ones — but 
there is only one type of chemistry: it goes on both inside our brains and bodies and also in our flasks 
and reactors, born from the ideas in our minds and the skill in our hands. We are not going to set 
ourselves up as moral judges in any way. We believe it is right to try and understand the world about 
us as best we can and to use that understanding creatively. This is what we want to share with 
you. 



Organic compounds 



Organic chemistry started as the chemistry of life, when that was thought to be different from the 
chemistry in the laboratory. Then it became the chemistry of carbon compounds, especially those 
found in coal. Now it is both. It is the chemistry of the compounds of carbon along with other ele- 
ments such as are found in living things and elsewhere. 




ll-c/'s-retinal 
absorbs light when we see 




serotonin 
human neurotransmitter 



We are going to give you 
structures of organic compounds 
in this chapter — otherwise it 
would be rather dull. If you do not 
understand the diagrams, do not 
worry. Explanation is on its way. 



1 ■ What is organic chemistry? 



The organic compounds available to us today are those present in living things and those formed 
over millions of years from dead things. In earlier times, the organic compounds known from nature 
were those in the 'essential oils' that could be distilled from plants and the alkaloids that could be 
extracted from crushed plants with acid. Menthol is a famous example of a flavouring compound 
from the essential oil of spearmint and ds-jasmone an example of a perfume distilled from jasmine 
flowers. 



You will be able to read towards the 
end of the book (Chapters 49-51) 
about the extraordinary chemistry that 
allows life to exist but this is known 
only from a modem cooperation 
between chemists and biologists. 





menthol 



c/s-jasmone 



MeO 




quinine 



Even in the sixteenth century one alkaloid was famous — quinine was extracted from the bark of 
the South American cinchona tree and used to treat fevers, especially malaria. The Jesuits who did 
this work (the remedy was known as 'Jesuit's bark') did not of course know what the structure of 
quinine was, but now we do. 

The main reservoir of chemicals available to the nineteenth century chemists was coal. Distil- 
lation of coal to give gas for lighting and heating (mainly hydrogen and carbon monoxide) also 
gave a brown tar rich in aromatic compounds such as benzene, pyridine, phenol, aniline, and 

thiophene. 

.OH /NH 2 jr. a 




benzene 







phenol 



aniline 



thiophene 



Phenol was used by Lister as an antiseptic in surgery and aniline became the basis for the dyestuffs 
industry. It was this that really started the search for new organic compounds made by chemists 
rather than by nature. A dyestuff of this kind — still available — is Bismarck Brown, which should tell 
you that much of this early work was done in Germany. 



H 2 N 




NH, 



,^' 




H 2 N 



'^, 




NH 2 



You can read about polymers and 
plastics in Chapter 52 and about fine 
chemicals throughout the book. 



CH3 (CH2)n CH3 

n = an enormous number 
length of molecule is n + 2 
carbon atoms 

CH 3 — (CH 2 )— CH 2 — CH 3 

n = an enormous number 
length of molecule is n + 3 
carbon atoms 



Bismarck Brown Y 

In the twentieth century oil overtook coal as the main source of bulk organic compounds so that 
simple hydrocarbons like methane (CH., 'natural gas') and propane (CH,CH 2 CH,, 'calor gas') 
became available for fuel. At the same time chemists began the search for new molecules from new 
sources such as fungi, corals, and bacteria and two organic chemical industries developed in paral- 
lel — 'bulk' and 'fine' chemicals. Bulk chemicals like paints and plastics are usually based on simple 
molecules produced in multitonne quantities while fine chemicals such as drugs, perfumes, and 
flavouring materials are produced in smaller quantities but much more profitably. 

At the time of writing there were about 16 million organic compounds known. How many more 
are possible? There is no limit (except the number of atoms in the universe). Imagine you've just 
made the longest hydrocarbon ever made — you just have to add another carbon atom and you've 
made another. This process can go on with any type of compound ad infinitum. 

But these millions of compounds are not just a long list of linear hydrocarbons; they embrace all 
kinds of molecules with amazingly varied properties. In this chapter we offer a selection. 



Organic compounds 



What do they look like? They may be crystalline solids, oils, 
waxes, plastics, elastics, mobile or volatile liquids, or gases. 
Familiar ones include white crystalline sugar, a cheap natural 
compound isolated from plants as hard white crystals when pure, 
and petrol, a mixture of colourless, volatile, flammable hydrocar- 
bons. Isooctane is a typical example and gives its name to the 
octane rating of petrol. 

The compounds need not lack colour. Indeed we can soon 
dream up a rainbow of organic compounds covering the whole 
spectrum, not to mention black and brown. In this table we have 
avoided dyestuffs and have chosen compounds as varied in struc- 
ture as possible. 



HO 1 





CH, 



CH 3 
CH 3 



H 2 



CH 3 
CH 



CH 3 



sucrose - ordinary sugar 

isolated from sugar cane 

or sugar beet 

white crystalline solid 



isooctane (2,3,5-trimethylpentane) 
a major constiuent of petrol 
volatile inflammable liquid 



s Colour 



red 



green 



blue 



purple 



Description 



Compound 



Structure 



dark red hexagonal plates 



3'-methoxybenzocycloheptatriene- 
2'-one 



MeO 




orange amber needles 



dichloro dicyano quinone (DDQ) 




yellow toxic yellow explosive gas diazomethane 



© 
CH 2 =N=N 



green prisms with a 
steel-blue lustre 



9-nitrosojulolidine 



deep blue liquid with a 
peppery smell 



deep blue gas condensing 
to a purple solid 



azulene 



nitroso trifluoromethane 




F'l 



Colour is not the only characteristic by which we recognize compounds. All too often it is their 
odour that lets us know they are around. There are some quite foul organic compounds too; the 
smell of the skunk is a mixture of two thiols — sulfur compounds containing SH groups. 

skunk spray contains: 



SH 





1 ■ What is organic chemistry? 



thioacetone 





X" 



trithioacetone; 
Freiburg was evacuated 
because of a smell from 
the distillation this compound 



HS SH 



HS 



But perhaps the worst aroma was that which caused the evacuation of the city of Freiburg in 1889. 
Attempts to make thioacetone by the cracking of trithioacetone gave rise to c an offensive smell which 
spread rapidly over a great area of the town causing fainting, vomiting and a panic evacuation°the 
laboratory work was abandoned'. 

It was perhaps foolhardy for workers at an Esso research station to repeat the experiment of crack- 
ing trithioacetone south of Oxford in 1967. Let them take up the story. 'Recently°we found ourselves 
with an odour problem beyond our worst expectations. During early experiments, a stopper jumped 
from a bottle of residues, and, although replaced at once, resulted in an immediate complaint of nau- 
sea and sickness from colleagues working in a building two hundred yards away. Two of our 
chemists who had done no more than investigate the cracking of minute amounts of trithioace- 
tone°found themselves the object of hostile stares in a restaurant and suffered the humiliation of 
having a waitress spray the area around them with a deodorant . The odours defied the expected 
effects of dilution since workers in the laboratory did not find the odours intolerable . . . and genu- 
inely denied responsibility since they were working in closed systems. To convince them otherwise, 
they were dispersed with other observers around the laboratory, at distances up to a quarter of a 
mile, and one drop of either acetone gem-dithiol or the mother liquors from crude trithioacetone 
crystallisations were placed on a watch glass in a fume cupboard. The odour was detected downwind 
in seconds.' 

There are two candidates for this dreadful smell — propane dithiol (called acetone gem-dithiol 
above) or 4-methyl-4-sulfanylpentan-2-one. It is unlikely that anyone else will be brave enough to 
resolve the controversy. 

Nasty smells have their uses. The natural gas piped to our homes contains small amounts of delib- 
erately added sulfur compounds such as ferf-butyl thiol (CH,),CSH. When we say small, we mean 
very small — humans can detect one part in 50 000 000 000 parts of natural gas. 

Other compounds have delightful odours. To redeem the honour of sulfur compounds we must 
cite the truffle which pigs can smell through a metre of soil and whose taste and smell is so delightful 
that truffles cost more than their weight in gold. Damascenones are responsible for the smell of roses. 
If you smell one drop you will be disappointed, as it smells rather like turpentine or camphor, but 
next morning you and the clothes you were wearing will smell powerfully of roses. Just like the com- 
pounds from trithioacetone, this smell develops on dilution. 

Humans are not the only creatures with a sense of smell. We can find mates using our eyes alone 
(though smell does play a part) but insects cannot do this. They are small in a crowded world and 
they find others of their own species and the opposite sex by smell. Most insects produce volatile 
compounds that can be picked up by a potential mate in incredibly weak concentrations. Only 1.5 
mg of serricornin, the sex pheromone of the cigarette beetle, could be isolated from 65 000 female 
beetles — so there isn't much in each beetle. Nevertheless, the slightest whiff of it causes the males to 
gather and attempt frenzied copulation. 

The sex pheromone of the Japanese beetle, also given off by the females, has been made by 
chemists. As little as 5 ug (micrograms, note!) was more effective than four virgin females in attract- 
ing the males. 



propane 
dithiol 



4-methyl-4- 

sulfanylpentan- 

2-one 



two candidates for 
the worst smell in the world 



no-one wants to find the winner! 



the divine smell 

of the black truffle 

comes from this compound 




damascenone - the smell of roses 





the sex pheromone of the cigarette beetle 
Lasioderma serricorne 



japonilure 

the sex pheromone of the Japanese beetle 
Popilia japonica 



The pheromone of the gypsy moth, disparlure, was identified from a few ug isolated from the 
moths and only 10 ug of synthetic material. As little as 2 x 10 g is active as a lure for the males in 
field tests. The three pheromones we have mentioned are available commercially for the specific 
trapping of these destructive insect pests. 



Organic compounds 



Don't suppose that the females always do all the work; both 
male and female olive flies produce pheromones that attract the 
other sex. The remarkable thing is that one mirror image of 
the molecule attracts the males while the other attracts the 
females! 





olean 

sex pheromone of the olive fly 
Bacrocera oieae 



this mirror image isomer 
attracts the males 





this mirror image isomer 
attracts the females 



the sex pheromone of the Gypsy moth 
Portheria dispar 



What about taste? Take the grapefruit. The main flavour comes from another sulfur compound 
and human beings can detect 2 x 1CT parts per billion of this compound. This is an almost unimag- 
inably small amount equal to 10~ mg per tonne or a drop, not in a bucket, but in a good-sized lake. 
Why evolution should have left us abnormally sensitive to grapefruit, we leave you to imagine. 

For a nasty taste, we should mention 'bittering agents', put into dangerous household substances 
like toilet cleaner to stop children eating them by accident. Notice that this complex organic com- 
pound is actually a salt — it has positively charged nitrogen and negatively charged oxygen atoms — 
and this makes it soluble in water. 

v poi oi *« »k*uH &# W OrtJ 

lllhll^jly ffMTTin' 1 ' E™? ll:HT U.I 

inning pwducb or bleach 



■..:■■! 




bitrex 

denatonium benzoate 

benzyldiethyl[(2,6-xylylcarbamoyl)methyl]ammonium benzoate 




"JMrftnlasKitfl A gallon 
«. -mil Hill, .loon ™w. 




flavouring principle of grapefruit 



CH 3 ^ ^OH 

alcohol 
(ethanol) 



Other organic compounds have strange effects on humans. Various 'drugs' such 
as alcohol and cocaine are taken in various ways to make people temporarily happy. 
They have their dangers. Too much alcohol leads to a lot of misery and any cocaine 
at all may make you a slave for life. 

Again, let's not forget other creatures. Cats seem to be able to go to sleep at any 
time and recently a compound was isolated from the cerebrospinal fluid of cats that makes them, or 
rats, or humans go off to sleep quickly. It is a surprisingly simple compound. 



a sleep-inducing fatty acid derivative 
c/s-9,10-octadecenoamide 




cocaine 
i addictive alkaloid 




This compound and disparlure are both derivatives of fatty 
acids, molecules that feature in many of the food problems people 
are so interested in now (and rightly so). Fatty acids in the diet are 
a popular preoccupation and the good and bad qualities of satu- 
rates, monounsaturates, and polyunsaturates are continually in 
the news. This too is organic chemistry. One of the latest mole- 
cules to be recognized as an anticancer agent in our diet is CLA 
(conjugated linoleic acid) in dairy products. 



11 



12 



10 




CLA (Conjugated Linoleic Acid) 

c/s-9-trans-ll conjugated linoleic acid 

dietary anticancer agent 



6 



1 ■ What is organic chemistry? 



HO 





Another fashionable molecule is resveratrole, which may ^-^ ^.OH 

be responsible for the beneficial effects of red wine in pre- 
venting heart disease. It is a quite different organic com- 
pound with two benzene rings and you can read about it in 
Chapter 51. 

For our third edible molecule we choose vitamin C. This is 
an essential factor in our diets — indeed, that is why it is called 
a vitamin. The disease scurvy, a degeneration of soft tissues, 
particularly in the mouth, from which sailors on long voyages 
like those of Columbus suffered, results if we don't have vitamin C. It also is a universal antioxidant, 
scavenging for rogue free radicals and so protecting us against cancer. Some people think an extra 
large intake protects us against the common cold, but this is not yet proved. 



Vitamin C (ascorbic acid) is a 
vitamin for primates, guinea-pigs, 
and fruit bats, but other mammals 
can make it for themselves. 



OH 



resveratrole from the skins of grapes 

is this the compound in red wine 
which helps to prevent heart disease? 




=1 Organic chemistry and industry 



vitamin C (ascorbic acid) 




monomers for polymer 
manufacture 



Vitamin C is manufactured on a huge scale by Roche, a Swiss company. All over the world there are 
chemistry-based companies making organic molecules on scales varying from a few kilograms to 
thousands of tonnes per year. This is good news for students of organic chemistry; there are lots of 
jobs around and it is an international job market. The scale of some of these operations of organic 
chemistry is almost incredible. The petrochemicals industry processes (and we use the products!) 
over 10 million litres of crude oil every day. Much of this is just burnt in vehicles as petrol or diesel, 
but some of it is purified or converted into organic compounds for use in the rest of the chemical 
industry. Multinational companies with thousands of employees such as Esso (Exxon) and Shell 
dominate this sector. 

Some simple compounds are made both from oil and from plants. The ethanol used as a starting 
material to make other compounds in industry is largely made by the catalytic hydration of ethylene 
from oil. But ethanol is also used as a fuel, particularly in Brazil where it is made by fermentation of 
sugar cane wastes. This fuel uses a waste product, saves on oil imports, and has improved the quality 
of the air in the very large Brazilian cities, Rio de Janeiro and Sao Paulo. 

Plastics and polymers take much of the production of the petro- 
chemical industry in the form of monomers such as styrene, acry- 
lates, and vinyl chloride. The products of this enormous industry are 
everything made of plastic including solid plastics for household 
goods and furniture, fibres for clothes (24 million tonnes per 
annum), elastic polymers for car tyres, light bubble-filled polymers 
for packing, and so on. Companies such as BASF, Dupont, Amoco, 
Monsanto, Laporte, Hoechst, and ICI are leaders here. Worldwide 
polymer production approaches 100 million tonnes per annum and 
PVC manufacture alone employs over 50 000 people to make over 20 
million tonnes per annum. 

The washing-up bowl is plastic too but the detergent you put in it belongs to another branch of 
the chemical industry — companies like Unilever (Britain) or Procter and Gamble (USA) which 
produce soap, detergent, cleaners, bleaches, 
polishes, and all the many essentials for the 
modern home. These products may be lemon 
and lavender scented but they too mostly come 
from the oil industry. Nowadays, most pro- 
ducts of this kind tell us, after a fashion, what is in 
them. Try this example — a well known brand of 
shaving gel along with the list of contents on the 
container: 

Does any of this make any sense? 




styrene 



-"% 





acrylates 



CI 



vinyl chloride 




Ingredients 

aqua, palmitic acid, 

triethanolamine, 

glycereth-26, isopentane, 

oleamide-DEA, oleth-2, 

stearic acid, isobutane, 

PEG-14M, parfum, 

allantoin, 

hydroxyethyl-cellulose, 

hydroxypropyl-cellulose, 

PEG-150 distearate, 

CI 42053, CI 47005 



Organic chemistry and industry 



It doesn't all make sense to us, but here is a possible interpretation. We certainly hope the book 
will set you on the path of understanding the sense (and the nonsense!) of this sort of thing. 



Ingredient 

aqua 


Chemical meaning 

water 


Purpose 

solvent 


palmitic acid 


CH 3 (CH 2 ) 14 C0 2 H 


acid, emulsifier 


triethanolamine 


N(CH 2 CH 2 0H) 3 


base 


glycereth-26 


glyceryl(0CH 2 CH 2 ) 26 0H 


surfactant 


isopentane 


(CH 3 ) 2 CHCH 2 CH 3 


propellant 


oleamide-DEA 


CH 3 (CH 2 ) 7 CH=CH(CH 2 ) 7 CONEt 2 




oleth-2 


0leyl(0CH 2 CH 2 ) 2 0H 


surfactant 


stearic acid 


CH 3 (CH 2 ) 16 C0 2 H 


acid, emulsifier 


isobutane 


(CH 3 ) 2 CHCH 3 


propellant 


PEG-14M 


polyoxyethylene glycol ester 


surfactant 


parfum 


perfume 




allantoin 


H 
H 2 N. ^N„ 

Y r~ m 


promotes healing in 
case you cut 
yourself while shaving 



allantoin 



H 



hydroxyethyl-cellulose 

hydroxypropyl-cellulose 

PEG-150 distearate 
CI 42053 
CI 47005 



cellulose fibre from wood pulp 
with -0CH 2 CH 2 0H groups added 

cellulose fibre from wood pulp 

with -0CH 2 CH(0H)CH 3 groups added 

polyoxyethylene glycol diester 

Fast Green FCF (see box) 

Quinoline Yellow (see box) 



gives body 

gives body 

surfactant 
green dye 
yellow dye 



The structures of two dyes 

Fast Green FCF and Quinoline Yellow are colours permitted to be used in foods and cosmetics and have the structures 
shown here. Quinoline Yellow is a mixture of isomeric sulfonic acids in the two rings shown. 



so,cP 




Fast Green FCF 



H00 2 S 




S0 2 OH 

Quinoline Yellow 



The particular acids, bases, surfactants, and so on are chosen to blend together in a smooth emul- 
sion when propelled from the can. The result should feel, smell, and look attractive and a greenish 
colour is considered clean and antiseptic by the customer. What the can actually says is this: 
'Superior lubricants within the gel prepare the skin for an exceptionally close, comfortable and effec- 
tive shave. It contains added moisturisers to help protect the skin from razor burn. Lightly 
fragranced.' 



8 



1 ■ What is organic chemistry? 



CN 




^CH 3 



Superglue bonds things together 

when this small molecule 

joins up with hundreds of its fellows 

in a polymerization reaction 



The formation of polymers is discussed 
in Chapter 52. 




Another oil-derived class of organic chemical business includes adhesives, sealants, coatings, and 
so on, with companies like Ciba-Geigy, Dow, Monsanto, and Laporte in the lead. Nowadays aircraft 
are glued together with epoxy-resins and you can glue almost anything with 'Superglue' a polymer of 
methyl cyanoacrylate. 

There is a big market for intense colours for dyeing cloth, colouring plastic and paper, painting 
walls, and so on. This is the dyestuffs and pigments industry and leaders here are companies like ICI 
and Akzo Nobel. ICI have a large stake in this aspect of the business, their paints turnover alone 
being £2 003 000 000 in 1995. 

The most famous dyestuff is probably indigo, an ancient dye that used to be isolated from plants 
but is now made chemically. It is the colour of blue jeans. More modern dyestuffs can be represented 
by ICI's benzodifuranones, which give fashionable red colours to synthetic fabrics like polyesters. 

We see one type of pigment around us all the time in the form of the colours on plastic bags. 
Among the best compounds for these are the metal complexes called phthalocyanines. Changing the 
metal (Cu and Fe are popular) at the centre and the halogens round the edge of these molecules 
changes the colour but blues and green predominate. The metal atom is not necessary for intense 
pigment colours — one new class of intense 'high performance' pigments in the orange-red range are 
the DPP (l,4-diketopyrrolo[3,4-c]pyrroles) series developed by Ciba-Geigy. Pigment Red 254 is 
used in paints and plastics. 




indigo 
the colour of blue jeans 



ICI's Dispersol 

benzodifuranone 

red dyes for polyester 




ICI's Monastral Green GNA 

a good green for plastic objects 



Ciba Geigy's Pigment Red 254 
an intense DPP pigment 



You can read in Chapter 7 why some 
compounds are coloured and others 
not. 



Colour photography starts with inorganic silver halides but they are carried on organic gelatin. 
Light acts on silver halides to give silver atoms that form the photographic image, but only in black 
and white. The colour in films like Kodachrome then comes from the coupling of two colourless 
organic compounds. One, usually an aromatic amine, is oxidized and couples with the other to give a 
coloured compound. 




NEt, 



light, silver 

*- 

photographic 

developer 



colourless 
aromatic amine 





OPh 



Q © 
S0 2 Na u 



magenta pigment from two 
NEt2 colourless compounds 




Na® S0 2 Q 



colourless cyclic amide 



Organic chemistry and industry 



That brings us to flavours and fragrances. Companies like International Flavours and Fragrances 
(USA) or Givaudan-Roure (Swiss) produce very big ranges of fine chemicals for the perfume, cos- 
metic, and food industries. Many of these will come from oil but others come from plant sources. A 
typical perfume will contain 5-10% fragrances in an ethanol/water (about 90:10) mixture. So the 
perfumery industry needs a very large amount of ethanol and, you might think, not much perfumery 
material. In fact, important fragrances like jasmine are produced on a > 10 000 tonnes per annum 
scale. The cost of a pure perfume ingredient like ds-jasmone, the main ingredient of jasmine, maybe 
several hundred pounds, dollars, or euros per gram. 




The world of perfumery 

Perfume chemists use extraordinary language to describe 
their achievements: 'Paco Rabanne pour homme was 
created to reproduce the effect of a summer walk in the 
open air among the hills of Provence: the smell of herbs, 
rosemary and thyme, and sparkling freshness with cool 
sea breezes mingling with warm soft Alpine air. To 



achieve the required effect, the perfumer blended 
herbaceous oils with woody accords and the synthetic 
aroma chemical dimethylheptanol which has a 
penetrating but indefinable freshness associated with 
open air or freshly washed linen'. (J. Ayres, Chemistry and 
Industry, 1988, 579) 



c/s-jasmone 

the main compound 

in jasmine perfume 




Chemists produce synthetic flavourings such as 'smoky bacon' and even 'chocolate'. Meaty 
flavours come from simple heterocycles such as alkyl pyrazines (present in coffee as well as roast 
meat) and furonol, originally found in pineapples. Compounds such as corylone and maltol give 
caramel and meaty flavours. Mixtures of these and other synthetic compounds can be 'tuned' to taste 
like many roasted foods from fresh bread to coffee and barbecued meat. 




an alkyl pyrazine 

from coffee and 

roast meat 



v 

AX 



furonol 
roast meat 




HO 



corylone 

caramel 

roasted taste 



*> 



maltol 

E-636 for cakes 

and biscuits 



Some flavouring compounds are also perfumes and may also be used as an intermediate in 
making other compounds. Two such large-scale flavouring compounds are vanillin (vanilla flavour 
as in ice cream) and menthol (mint flavour) both manufactured on a large scale and with many 
uses. 



van "" n rw n 
found in vanilla pods; l ' n 3 u 

manufactured 

on a large scale 

HO 





menthol 

extracted from mint; 

25% of the world's supply 

manufactured 




Food chemistry includes much larger-scale items than flavours. Sweeteners such as sugar itself are 
isolated from plants on an enormous scale. Sugar's structure appeared a few pages back. Other 
sweeteners such as saccharin (discovered in 1879!) and aspartame (1965) are made on a sizeable 
scale. Aspartame is a compound of two of the natural amino acids present in all living things and is 
made by Monsanto on a large scale (over 10 000 tonnes per annum). 



,C0 2 H 



methyl ester of 
phenylalanine 




is made from 
two amino acids - 



H,N 



aspartame ('NutraSweet') 
200 x sweeter than sugar 





10 



1 ■ What is organic chemistry? 



The pharmaceutical businesses produce drugs and medicinal products of many kinds. One of the 
great revolutions of modern life has been the expectation that humans will survive diseases because 
of a treatment designed to deal specifically with that disease. The most successful drug ever is raniti- 
dine (Zantac), the Glaxo-Wellcome ulcer treatment, and one of the fastest-growing is Pfizer's silde- 
nafil (Viagra). 'Success' refers both to human health and to profit! 

You will know people (probably older men) who are 'on (3-blockers'. These are com- 
pounds designed to block the effects of adrenaline (epinephrine) on the heart and hence to 
prevent heart disease. One of the best is Zeneca's tenormin. Preventing high blood pressure also pre- 
vents heart disease and certain specific enzyme inhibitors (called 'ACE-inhibitors') such as 
Squibb's captopril work in this way. These are drugs that imitate substances naturally present in the 
body. 

The treatment of infectious diseases relies on antibiotics such as the penicillins to prevent bacteria 
from multiplying. One of the most successful of these is Smith Kline Beecham's amoxycillin. The 
four-membered ring at the heart of the molecule is the '(3-lactam'. 



Me 2 N 




EtO 



N NHMe 

H 



Glaxo-Wellcome's ranitidine 

the most successful drug to date 

world wide sales peaked >£1, 000, 000, 000 per annum 




Me 



Pfizer's sildenafil (Viagra) 

three million satisfied customers in 1998 




Zeneca's tenormin 

cardioselective pi-blocker 

for treatment and prevention 

of heart disease 



C0 2 H 



Squibb's captopril 
specific enzyme inhibitor 

for treatment and 
prevention of hypertension 




SmithKline Beecham's amoxycillin 

P-lactam antibiotic 
for treatment of bacterial infections 



C0 2 H 



We cannot maintain our present high density of population in the developed world, nor deal with 
malnutrition in the developing world unless we preserve our food supply from attacks by insects and 
fungi and from competition by weeds. The world market for agrochemicals is over £10 000 000 000 
per annum divided roughly equally between herbicides, fungicides, and insecticides. 

At the moment we hold our own by the use of agrochemicals: companies such as Rhone- 
Poulenc, Zeneca, BASF, Schering-Plough, and Dow produce compounds of remarkable and specific 
activity. The most famous modern insecticides are modelled on the natural pyrethrins, stabilized 
against degradation by sunlight by chemical modification (see coloured portions of decamethrin) 
and targeted to specific insects on specific crops in cooperation with biologists. Decamethrin has a 
safety factor of >10#000 for mustard beetles over mammals, can be applied at only 10 grams per 
hectare (about one level tablespoon per football pitch), and leaves no significant environmental 
residue. 




a natural pyrethin 
from pyrethrum - daisy-like flowers from East Africa 



decamethrin 
a modified pyrethrin - more active and stable in sunlight 



Organic chemistry and the periodic table 



11 



As you learn more chemistry, you will appreciate how remarkable it is that Nature should pro- 
duce three-membered rings and that chemists should use them in bulk compounds to be sprayed on 
crops in fields. Even more remarkable in some ways is the new generation of fungicides based on a 
five-membered ring containing three nitrogen atoms — the triazole ring. These compounds inhibit 
an enzyme present in fungi but not in plants or animals. 

One fungus (potato blight) caused the Irish potato famine of the nineteenth century and the vari- 
ous blights, blotches, rots, rusts, smuts, and mildews can overwhelm any crop in a short time. 
Especially now that so much is grown in Western Europe in winter, fungal diseases are a real 
threat. 



,C0 2 Me 




benomyl 

a fungicide which controls 

many plant diseases 




propiconazole 
a triazole fungicide 



You will have noticed that some of these companies have fingers in many pies. These companies, 
or groups as they should be called, are the real giants of organic chemistry. Rhone-Poulenc, the 
French group which includes pharmaceuticals (Rhone-Poulenc-Rorer), animal health, agrochemi- 
cals, chemicals, fibres, and polymers, had sales of about 90 billion French Francs in 1996. Dow, the 
US group which includes chemicals, plastics, hydrocarbons, and other bulk chemicals, had sales of 
about 20 billion US dollars in 1996. 



Organic chemistry and the periodic table 

All the compounds we have shown you are built up on hydrocarbon (carbon and hydrogen) skele- 
tons. Most have oxygen and/or nitrogen as well; some have sulfur and some phosphorus. These are 
the main elements of organic chemistry but another way the science has developed is an exploration 
of (some would say take-over bid for) the rest of the periodic table. Some of our compounds also had 
fluorine, sodium, copper, chlorine, and bromine. The organic chemistry of silicon, boron, lithium, 
the halogens (F, CI, Br, and I), tin, copper, and palladium has been particularly well studied and 
these elements commonly form part of organic reagents used in the laboratory. They will crop up 
throughout this book. These 'lesser' elements appear in many important reagents, which are used in 
organic chemical laboratories all over the world. Butyllithium, trimethylsilyl chloride, tributyltin 
hydride, and dimethylcopper lithium are good examples. 

The halogens also appear in many life-saving drugs. The recently discovered antiviral com- 
pounds, such as fialuridine (which contains both F and I, as well as N and O), are essential for the 
fight against HIV and AIDS. They are modelled on natural compounds from nucleic acids. The 
naturally occurring cytotoxic (antitumour) agent halomon, extracted from red algae, contains Br 
and CI. 



CH 



CH 3 

■> 

CH 3 



- 



C4H9 

H q — Sn- 

/ 
C4H9 



'4 n 9 



BuLi 

butyllithium 



Me 3 SiCI 
trimethylsilyl chloride 



Bu 3 SnH 
tributyltin hydride 



CH 3 

))u Li® 
CH 3 

Me 2 CuLi 
dimethylcopper lithium 




halomon 
naturally occurring 
antitumour agent 



fialuridine 
antiviral compound 



Another definition of organic chemistry would use the periodic table. The key elements in 
organic chemistry are of course C, H, N, and O, but also important are the halogens (F, CI. Br, I), 



12 



1 ■ What is organic chemistry? 



p-block elements such as Si, S, and P, metals such as Li, Pd, Cu, and Hg, and many more. We can 
construct an organic chemist's periodic table with the most important elements emphasized: 



You will certainly know something 
about the periodic table from your 
previous studies of inorganic 
chemistry. A basic knowledge of 
the groups, which elements are 
metals, and roughly where the 
elements in ourtable appearwill 
be helpful to you. 



ao 



the organic chemist's 
periodic table 



13 14 



15 



16 



17 



ogkdcdq 

00000000000000 oa 



Na M 



10 11 



ooooooooo0ooo0ooo 
oooooooooooqooooo 

So where does inorganic chemistry end and organic chemistry begin? Would you say that 
the antiviral compound foscarnet was organic? It is a compound of carbon with the formula 
CPCXNa, but is has no C-H bonds. And what about the important reagent tetrakis triphenyl phos- 
phine palladium? It has lots of hydrocarbon — twelve benzene rings in fact — but the benzene rings are 
all joined to phosphorus atoms that are arranged in a square around the central palladium atom, so 
the molecule is held together by C-P and P-Pd bonds, not by a hydrocarbon skeleton. Although it has 
the very organic-looking formula C 79 H fin P 4 Pd, many people would say it is inorganic. But is it? 




tetrakis 

triphenylphosphine 

palladium 

[(CeHshPkPd 
(Ph 3 P) 4 Pd 



GO 





7 Y 



© 



foscarnet - antiviral agent 



The answer is that we don't know and we don't care. It is important these days to realize that 
strict boundaries between traditional disciplines are undesirable and meaningless. Chemistry 
continues across the old boundaries between organic chemistry and inorganic chemistry on the one 
side and organic chemistry and biochemistry on the other. Be glad that the boundaries are indistinct 
as that means the chemistry is all the richer. This lovely molecule (Ph,P).Pd belongs to chemistry. 



Organic chemistry and this book 



We have told you about organic chemistry's history, the types of compounds it concerns itself with, the 
things it makes, and the elements it uses. Organic chemistry today is the study of the structure and reac- 
tions of compounds in nature of compounds, in the fossil reserves such as coal and oil, and of those 
compounds that can be made from them. These compounds will usually be constructed with a hydro- 
carbon framework but will also often have atoms such as O, N, S, P, Si, B, halogens, and metals attached 
to them. Organic chemistry is used in the making of plastics, paints, dyestuffs, clothes, foodstuffs, 
human and veterinary medicines, agrochemicals, and many other things. Now we can summarize all of 
these in a different way. 

# The main components of organic chemistry as a discipline are these 

• Structure determination — how to find out the structures of new compounds 
even if they are available only in invisibly small amounts 

• Theoretical organic chemistry — how to understand those structures in terms 
of atoms and the electrons that bind them together 

• Reaction mechanisms — how to find out how these molecules react with each 
other and how to predict their reactions 

• Synthesis — how to design new molecules — and then make them 

• Biological chemistry — how to find out what Nature does and how the 
structures of biologically active molecules are related to what they do 

This book is about all these things. It tells you about the structures of organic molecules and the 
reasons behind them. It tells you about the shapes of those molecules and how the shape relates to 
their function, especially in the context of biology. It tells you how those structures and shapes are 
discovered. It tells you about the reactions the molecules undergo and, more importantly, how and 
why they behave in the way they do. It tells you about nature and about industry. It tells you how 
molecules are made and how you too can think about making molecules. 

We said 'it tells' in that last paragraph. Maybe we should have said 'we tell' because we want to 
speak to you through our words so that you can see how we think about organic chemistry and to 
encourage you to develop your own ideas. We expect you to notice that four people have written this 
book and that they don't all think or write in the same way. That is as it should be. Organic chemistry 
is too big and important a subject to be restricted by dogmatic rules. Different chemists think in dif- 
ferent ways about many aspects of organic chemistry and in many cases it is not yet possible to be 
sure who is right. 

We may refer to the history of chemistry from time to time but we are usually going to tell you about 
organic chemistry as it is now. We will develop the ideas slowly, from simple and fundamental ones 
using small molecules to complex ideas and large molecules. We promise one thing. We are not going 
to pull the wool over your eyes by making things artificially simple and avoiding the awkward ques- 
tions. We aim to be honest and share both our delight in good complete explanations and our puzzle- 
ment at inadequate ones. So how are we going to do this? The book starts with a series of chapters on 
the structures and reactions of simple molecules. You will meet the way structures are determined and 
the theory that explains those structures. It is vital that you realize that theory is used to explain what is 
known by experiment and only then to predict what is unknown. You will meet mechanisms — the 
dynamic language used by chemists to talk about reactions — and of course some reactions. 



14 1 ■ Organic chemistry and this book 



The book starts with an introductory section of four chapters: 

1 What is organic chemistry? 

2 Organic structures 

3 Determining organic structures 

4 Structure of molecules 

In Chapter 2 you will look at the way in which we are going to present diagrams of molecules 
on the printed page. Organic chemistry is a visual, three-dimensional subject and the way you 
draw molecules shows how you think about them. We want you too to draw molecules in the best way 
available now. It is just as easy to draw them well as to draw them in an old-fashioned inaccurate way. 

Then in Chapter 3, before we come to the theory of molecular structure, we shall introduce you to 
the experimental techniques of finding out about molecular structure. This means studying the 
interactions between molecules and radiation by spectroscopy — using the whole electromagnetic 
spectrum from X-rays to radio waves. Only then, in Chapter 4, will we go behind the scenes and look 
at the theories of why atoms combine in the ways they do. Experiment comes before theory. The 
spectroscopic methods of Chapter 3 will still be telling the truth in a hundred years time, but the the- 
ories of Chapter 4 will look quite dated by then. 

We could have titled those three chapters: 

2 What shapes do organic molecules have? 

3 How do we know they have those shapes? 

4 Why do they have those shapes? 

You need to have a grasp of the answers to these three questions before you start the study of 
organic reactions. That is exactly what happens next. We introduce organic reaction mechanisms in 
Chapter 5. Any kind of chemistry studies reactions — the transformations of molecules into other 
molecules. The dynamic process by which this happens is called mechanism and is the language of 
organic chemistry. We want you to start learning and using this language straight away so in Chapter 
6 we apply it to one important class of reaction. This section is: 

5 Organic reactions 

6 Nucleophilic addition to the carbonyl group 

Chapter 6 reveals how we are going to subdivide organic chemistry. We shall use a mechanistic 
classification rather than a structural classification and explain one type of reaction rather than one 
type of compound in each chapter. In the rest of the book most of the chapters describe types of reac- 
tion in a mechanistic way. Here is a selection. 

9 Using organometallic reagents to make C-C bonds 

17 Nucleophilic substitution at saturated carbon 

20 Electrophilic addition to alkenes 

22 Electrophilic aromatic substitution 

29 Conjugate Michael addition of enolates 

39 Radicals 

Interspersed with these chapters are others on physical aspects, organic synthesis, stereochem- 
istry, structural determination, and biological chemistry as all these topics are important parts of 
organic chemistry. 

'Connections' section 

Chemistry is not a linear subject! It is impossible simply to start at the beginning and work through 
to the end, introducing one new topic at a time, because chemistry is a network of interconnecting 
ideas. But, unfortunately, a book is, by nature, a beginning-to-end sort of thing. We have arranged 
the chapters in a progression of difficulty as far as is possible, but to help you find your way around 



Boxes and margin notes 



15 



we have included at the beginning of each chapter a 'Connections' section. This tells you three things 
divided among three columns: 

(a) what you should be familiar with before reading the chapter — in other words, which previous 
chapters relate directly to the material within the chapter ('Building on' column) 

(b) a guide to what you will find within the chapter ('Arriving at' column) 

(c) which chapters later in the book fill out and expand the material in the chapter ('Looking 
forward to' column) 

The first time you read a chapter, you should really make sure you have read any chapter mentioned 
under (a). When you become more familiar with the book you will find that the links highlighted in 
(a) and (c) will help you see how chemistry interconnects with itself. 



Boxes and margin notes 



The other things you should look out for are the margin notes and boxes. There are four sorts, and 
they have all appeared at least once in this chapter. 



Heading 

The most important looks like this. Anything in this sort of box is very 
important — a key concept or a summary. It's the sort of thing you would do well to 
hold in your mind as you read or to note down as you learn. 



Heading 

Boxes like this will contain additional examples, amusing 
background information, and similar interesting, but 
inessential, material. The first time you read a chapter, 



you might want to miss out this sort of box, and only read 
them later on to flesh out some of the main themes of the 
chapter. 



End- of- chapter problems 

You can't learn organic chemistry — there's just too much of it. You can learn trivial things 
like the names of compounds but that doesn't help you understand the principles behind the 
subject. You have to understand the principles because the only way to tackle organic chemistry 
is to learn to work it out. That is why we have provided end-of-chapter problems. They are 
to help you discover if you have understood the material presented in each chapter. In general, 
the 10-15 problems at the end of each chapter start easy and get more difficult. They come 
in two sorts. The first, generally shorter and easier, allow you to revise the material in that chap- 
ter. The second asks you to extend your understanding of the material into areas not covered 
by the chapter. In the later chapters this second sort will probably revise material from previous 
chapters. 

If a chapter is about a certain type of organic reaction, say elimination reactions (Chapter 19), the 
chapter itself will describe the various ways ('mechanisms') by which the reaction can occur and it 
will give definitive examples of each mechanism. In Chapter 19 there are three mechanisms and 
about 65 examples altogether. You might think that this is rather a lot but there are in fact millions of 
examples known of these three mechanisms and Chapter 19 only scrapes the surface. Even if you 
totally comprehended the chapter at a first reading, you could not be confident of your understand- 
ing about elimination reactions. There are 13 end-of-chapter problems for Chapter 19. The first 
three ask you to interpret reactions given but not explained in the chapter. This checks that you can 
use the ideas in familiar situations. The next few problems develop specific ideas from the chapter 
concerned with why one compound does one reaction while a similar one behaves quite differently. 



Sometimes the main text of the 
book needs clarification or 
expansion, and this sort of margin 
note will contain such little extras 
to help you understand difficult 
points. It will also remind you of 
things from elsewhere in the book 
that illuminate what is being 
discussed. You would do well to 
read these notes the first time 
you read the chapter, though 
later, as the ideas become more 
familiar, you might choose to skip 
them. 



This sort of margin note will mainly 
contain cross-references to other 
parts of the book as a further aid to 
navigation. You will find an example 
on p. 000. 



16 



1 ■ Organic chemistry and this book 



Finally there are some more challenging problems asking you to extend the ideas to unfamiliar 
molecules. 

The end-of-chapter problems should set you on your way but they are not the end of the journey 
to understanding. You are probably reading this text as part of a university course and you should 
find out what kind of examination problems your university uses and practise them too. Your tutor 
will be able to advise you on suitable problems for each stage of your development. 

The solutions manual 

The problems would be of little use to you if you could not check your answers. For the 
maximum benefit, you need to tackle some or all of the problems as soon as you have finished 
each chapter without looking at the answers. Then you need to compare your suggestions with 
ours. You can do this with the solutions manual (Organic Chemistry: Solutions Manual, Oxford 
University Press, 2000). Each problem is discussed in some detail. The purpose of the problem 
is first stated or explained. Then, if the problem is a simple one, the answer is given. If the prob- 
lem is more complex, a discussion of possible answers follows with some comments on the value 
of each. There may be a reference to the source of the problem so that you can read further if you 
wish. 



Colour 

You will already have noticed something unusual about this book: almost all of the chemical struc- 
tures are shown in red. This is quite intentional: emphatic red underlines the message that structures 
are more important than words in organic chemistry. But sometimes small parts of structures are in 
other colours: here are two examples from p. 000, where we were talking about organic compounds 
containing elements other than C and H. 



fialuridine 
antiviral compound 





Halomon 
naturally occurring antitumour agent 



Why are the atom labels black? Because we wanted them to stand out from the rest of the 
molecule. In general you will see black used to highlight important details of a molecule — they may 
be the groups taking part in a reaction, or something that has changed as a result of the reaction, as in 
these examples from Chapters 9 and 12. 




MgBr 




1. EtMgBr 

2. H 3 + HO 



i=^ \A^ 



2. H\ H 2 n ew C-C bond 

We shall often use black to emphasize 'curly arrows', devices that show the movement of elec- 
trons, and whose use you will learn about in Chapter 5. Here is an example from Chapter 10: notice 
black also helps the '+' and '-' charges to stand out. 






CN 



Me 




Colour 



17 



Occasionally, we shall use other colours such as green, or even orange, yellow, or brown, to high- 
light points of secondary importance. This example is part of a reaction taken from Chapter 19: we 
want to show that a molecule of water (H z O) is formed. The green atoms show where the water 
comes from. Notice black curly arrows and a new black bond. 

new C=C 



OH 



H^^H 



double bond 




^S^ 



H 




■~^Q 



"To 



H,0 



Other colours come in when things get more complicated — in this Chapter 24 example, we want 
to show a reaction happening at the black group in the presence of the yellow H (which, as you will 
see in Chapter 9, also reacts) and also in the presence of the green 'protecting' groups, one of the 
topics of Chapter 24. 

,Ph ^Ph 



Me ° 2c yy^Y ^^i ho^^y^"^ 

And, in Chapter 16, colour helps us highlight the difference between carbon atoms carrying four 
different groups and those with only three different groups. The message is: if you see something in a 
colour other than red, take special note — the colour is there for a reason. 



4 H NH 3 



amino acids 
are chiral 



3 R 



X, 



C0 2 H 2 



3 H 



H / N " 2 



C0 2 H 2 



except glycine - 
plane of paper is a 
plane of symmetry 
through C, N, and C0 2 H 



That is all we shall say in the way of introduction. On the next page the real chemistry starts, and 
our intention is to help you to learn real chemistry, and to enjoy it. 



Organic structures 



2 



Connections 

Building on: 

• This chapter does not depend on 
Chapter 1 



Leading to: 

• The diagrams used in the rest of the book 

• Why we use these particular diagrams 

• How organic chemists name 
molecules in writing and in speech 

• What is the skeleton of an organic 
molecule 

• What is a functional group 

• Some abbreviations used by all organic 
chemists 

• Drawing organic molecules realistically 
in an easily understood style 



Looking forward to: 

• Ascertaining molecular structure 
spectroscopically ch3 

• What determines a molecule's 
structure ch4 



There are over 100 elements in the periodic table. Many molecules contain well over 100 atoms — 
palytoxin, for example (a naturally occurring compound with potential anticancer activity) contains 
129 carbon atoms, 221 hydrogen atoms, 54 oxygen atoms, and 3 nitrogen atoms. It's easy to see how 
chemical structures can display enormous variety, providing enough molecules to build even the 
most complicated living creatures. But how can we understand what seems like a recipe for confu- 
sion? Faced with the collection of atoms we call a molecule, how can we make sense of what we see? 
This chapter will teach you how to interpret organic structures. It will also teach you how to draw 
organic molecules in a way that conveys all the necessary information and none of the superfluous. 

OH 
OH '" 



Palytoxin was isolated in 1971 in 
Hawaii from Limu make o Hane ('deadly 
seaweed of Hana') which had been 
used to poison spear points. It is one of 
the most toxic compounds known 
requiring only about 0.15 microgram 
per kilogram for death by injection. The 
complicated structure was determined 
a few years later. 




palytoxin 



20 



2 ■ Organic structures 



Hydrocarbon frameworks and functional groups 

As we explained in Chapter 1, organic chemistry is the study of compounds that contain carbon. Nearly 
all organic compounds also contain hydrogen; most also contain oxygen, nitrogen, or other elements. 
Organic chemistry concerns itself with the way in which these atoms are bonded together into stable 
molecular structures, and the way in which these structures change in the course of chemical reactions. 
Some molecular structures are shown below. These molecules are all amino acids, the con- 
stituents of proteins. Look at the number of carbon atoms in each molecule and the way they are 
bonded together. Even within this small class of molecules there's great variety — glycine and alanine 
have only two or three carbon atoms; phenylalanine has nine. 



We shall return to amino acids as 
examples several times in this chapter, 
but we shall leave detailed discussions 
about their chemistry till Chapters 24 
and 49, when we look at the way in 
which they polymerize to form peptides 
and proteins. 



H NH 2 


H NH 2 

\/ 


N^ **C< H NH 2 


/C OH 

h Nr 

II 


C /OH 

ch 3 ^ Nr 

II 


II 1 \ / 2 
h^N^N^N^ " 








1 H /X H 'I 
H H H 



Lysine has a chain of atoms; tryptophan has rings. 



H H H H H NH 2 

\/ v/ V 

/C /C /C /OH 

H 2 ivr ^c^ Nr ^c 

/\ /\ || 

H H H H o 



H 

\ 



H 



N y» 

c y \ / 

H-^„/ ^9 ,9 H NH 2 

^c \ // \/ 

\\ yC— C /C ^OH 

c ^ 

H \ 



H 



H H 



In methionine the atoms are arranged in a single chain; in leucine the chain is branched. In proline, 
the chain bends back on itself to form a ring. 



H 3 C. 



H H H NH, 

\ / \ / 

/\ II 

H H ii 



OH 



H.I 



CH 



H 3 C 



H NH 2 

/\ II 

H H 



\ 



/ 



C — N 



OH 



/ 



/ X C 
H A 
H H 



V 



/ "V 



OH 



Yet all of these molecules have similar properties — they are all soluble in water, they are all both 
acidic and basic (amphoteric), they can all be joined with other amino acids to form proteins. This is 
because the chemistry of organic molecules depends much less on the number or the arrangement of 
carbon or hydrogen atoms than on the other types of atoms (O, N, S, P, Si...) in the molecule. We 
call parts of molecules containing small collections of these other atoms functional groups, simply 
because they are groups of atoms that determine the way the molecule works. All amino acids con- 
tain two functional groups: an amino (NH2 or NH) group and a carboxylic acid (CO2H) group 
(some contain other functional groups as well). 



The functional groups determine the way the molecule works both chemically 
and biologically. 



H NH 2 
\/ 
C /OH 

h 3 <t Nr 

II 



H H H H H NH, 

\ / \ / \ / 

h^N^N^N;' 

/\ /\ II 

H H H H 


alanine 


lysine 


contains just the 


has an additional 


and carboxylic acid 


amino group 


functional groups 





.OH 



H 3 C, 



H H H NH, 

/\ 11 

H H 

methionine 

also has a sulfide 
functional group 



OH 



Drawing molecules 



21 



That isn't to say the carbon atoms aren't important; they just play quite a different role from those 
of the oxygen, nitrogen, and other atoms they are attached to. We can consider the chains and rings 
of carbon atoms we find in molecules as their skeletons, which support the functional groups and 
allow them to take part in chemical interactions, much as your skeleton supports your internal 
organs so they can interact with one another and work properly. 

The hydrocarbon framework is made up of chains and rings of carbon atoms, and 
it acts as a support for the functional groups. 



H H H H H H 

\ / \ / \ / 

/\ /\ 

H H H H 



a chain 



H H 

H \/ H 

H-VN/-H 



H— C. JC— 

/ Nr^\ 

H /\ H 
H H 



a ring 



H 



H H H H H H 

\ / \ / \ / 
/ \ ^ \ / \ 

/\ /\ 

H — C H H C— H 

/\ /\ 

H H H H 

a branched chain 



We will see later how the interpretation of organic structures as hydrocarbon frameworks sup- 
porting functional groups helps us to understand and rationalize the reactions of organic molecules. 
It also helps us to devise simple, clear ways of representing molecules on paper. You saw in Chapter 1 
how we represented molecules on paper, and in the next section we shall teach you ways to draw 
(and ways not to draw) molecules — the handwriting of chemistry. This section is extremely impor- 
tant, because it will teach you how to communicate chemistry, clearly and simply, throughout your 
life as a chemist. 



Organic skeletons 

Organic molecules left to 
decompose for millions of years in 
the absence of light and oxygen 
become literally carbon 
skeletons — crude oil, for example, 
is a mixture of molecules 
consisting of nothing but carbon 
and hydrogen, while coal consists 
of little else but carbon. Although 
the molecules in coal and oil differ 
widely in chemical structure, they 
have one thing in common: no 
functional groups! Many are very 
unreactive: about the only 
chemical reaction they can take 
part in is combustion, which, in 
comparison to most reactions that 
take place in chemical laboratories 
or in living systems, is an 
extremely violent process. In 
Chapter 5 we will start to look at 
the way that functional groups 
direct the chemical reactions of a 
molecule. 



Drawing molecules 

Be realistic 

Below is another organic structure — again, you may be familiar with the molecule it represents; it is 
a fatty acid commonly called linoleic acid. 



HHHH HH HHHHHHHH 

\/ \/ \/ \/ \/ \/ \/ 

H3C \ c / c \ c / c \ c=c / c \ c=c /%./ c \ c /%/C^ 

/\ /\ / \ / \ /\ /\ /\ || 

HHHHH HH HHHHHHH q 

linoleic acid 
We could also depict linoleic acid as 

CH3CH2CH2CH2CH— CHCH2CH— CHCH2CH2CH2CH2CH2CH2CH2CO2H 

linoleic acid 
or as 



,0H 



carboxylic acid 
functional group 



H 



HHHH H HHHHHHH 

I I I I I I I I I I I I 

H — C — C — C — C — C — C=C — C — C=C — C — C — C — C — C — C — C — C0 2 H 

I I I I I I I I I I I I I I I I I 
HHHHHHHHHHHHHHHHH 

linoleic acid 

You may well have seen diagrams like these last two in older books — they used to be easy to print (in 
the days before computers) because all the atoms were in a line and all the angles were 90°. But are 
they realistic? We will consider ways of determining the shapes and structures of molecules in more 
detail in Chapter 3, but the picture below shows the structure of linoleic acid determined by X-ray 
crystallography. 



Three fatty acid molecules and one 
glycerol molecule combine to form the 
fats that store energy in our bodies and 
are used to construct the membranes 
around our cells. This particular fatty 
acid, linoleic acid, cannot be 
manufactured in the human body, and 
is an essential part of a healthy diet 
found, for example, in sunflower oil. 

Fatty acids differ in the length of their 
chains of carbon atoms, yet they have 
very similar chemical properties 
because they all contain the carboxylic 
acid functional group. We shall come 
back to fatty acids in Chapter 49. 

H OH 

\/ 

/\ /\ 

HHHH 

glycerol 



X-ray crystallography discovers 
the structures of molecules by 
observing the way X-rays bounce 
off atoms in crystalline solids. It 
gives clear diagrams with the 
atoms marked a circles and the 
bonds as rods joining them 
together. 



22 



2 ■ Organic structures 




You can see that the chain of carbon atoms is not linear, but a zig-zag. Although our diagram is just a 
two-dimensional representation of this three-dimensional structure, it seems reasonable to draw it 
as a zig-zag too. 



HHHH HH HHHHHHHH 

\/ \/ \/ \/ \/ \/ \/ 

HqC. j-C- ^C> j^C^ J-C- ^C- j-C- j^C- jfOH 

Nj^ N:^ ^c=c^ ^c=c^ Ns^ ^c^ N:^ Ns^ 

/\ /\ / \ / \ /\ /\ /\ II 

HHHHH HH HHHHHHH q 

Hnoleic acid 
This gives us our first guideline for drawing organic structures. 



• Guideline 1 

Draw chains of atoms as zig-zags 

Realism of course has its limits — the X-ray structure shows that the linoleic acid molecule is in fact 
slightly bent in the vicinity of the double bonds; we have taken the liberty of drawing it as a 'straight 
zig-zag'. Similarly, close inspection of crystal structures like this reveals that the angle of the zig-zag is 
about 109° when the carbon atom is not part of a double bond and 120° when it is. The 109° angle is 
the 'tetrahedral angle', the angle between two vertices of a tetrahedron when viewed from its centre. 
In Chapter 4 we shall look at why carbon atoms take up this particular arrangement of bonds. Our 
realistic drawing is a projection of a three-dimensional structure onto flat paper so we have to com- 
promise. 

Be economical 

When we draw organic structures we try to be as realistic as we can be without putting in superfluous 
detail. Look at these three pictures. 




12 3 

(1) is immediately recognizable as Leonardo da Vinci's Mona Lisa. You may not recognize (2) — it's 
also Leonardo da Vinci's Mona Lisa — this time viewed from above. The frame is very ornate, but the 
picture tells us as much about the painting as our rejected linear and 90° angle diagrams did about 



Drawing molecules 



23 



our fatty acid. They're both correct — in their way — but sadly useless. What we need when we draw 
molecules is the equivalent of (3). It gets across the idea of the original, and includes all the detail 
necessary for us to recognize what it's a picture of, and leaves out the rest. And it was quick to draw — 
this picture was drawn in less than 10 minutes: we haven't got time to produce great works of 
art! 

Because functional groups are the key to the chemistry of molecules, clear diagrams must empha- 
size the functional groups, and let the hydrocarbon framework fade into the background. Compare 
the diagrams below: 



HHHH HH HHHHHHHH 

\/ \/ \/ \/ \/ \/ \/ 

/\ /\ / \ / \ /\ /\ /\ | 

HHHHH HH HHHHHHH { 

linoleic acid 



,OH 




linoleic acid 



The second structure is the way that most organic chemists would draw linoleic acid. Notice how the 
important carboxylic acid functional group stands out clearly and is no longer cluttered by all those 
Cs and Hs. The zig-zag pattern of the chain is much clearer too. And this structure is much quicker 
to draw than any of the previous ones! 

To get this diagram from the one above we've done two things. Firstly, we've got rid of all the 
hydrogen atoms attached to carbon atoms, along with the bonds joining them to the carbon atoms. 
Even without drawing the hydrogen atoms we know they're there — we assume that any carbon atom 
that doesn't appear to have its potential for four bonds satisfied is also attached to the appropriate 
number of hydrogen atoms. Secondly, we've rubbed out all the Cs representing carbon atoms. We're 
left with a zig-zag line, and we assume that every kink in the line represents a carbon atom, as does 
the end of the line. 



the end of the line 
represents a C atom 



every kink in 

the chain represents 

a C atom 




this C atom must 
also carry 3 H atoms 
because only 1 bond 
is shown 



these C atoms must 

also carry 1 H atom 

because only 3 bonds 

are shown for each atom 



these C atoms must 

also carry 2 H atoms 

because only 2 bonds 

are shown for each atom 



this H is shown 
because it is 
attached to an 
atom other than C 



all four bonds 
are shown to this 
C atom, so no 
H atoms are 
implied 



We can turn these two simplifications into two more guidelines for drawing organic structures. 

• Guideline 2 

Miss out the Hs attached to carbon atoms, along with the C-H bonds 
(unless there is a good reason not to) 

• Guideline 3 

Miss out the capital Cs representing carbon atoms 
(unless there is a good reason not to) 



What is 'a good reason not to'? 
One is if the C or H is part of a 
functional group. Another is if the 
C or H needs to be highlighted in 
some way, for example, because 
it's taking part in a reaction. Don't 
be too rigid about these 
guidelines: they're not rules. 
Better is just to learn by example 
(you'll find plenty in this book): if it 
helps clarify, put it in; if it clutters 
and confuses, leave it out. One 
thing you must remember, 
though: if you write a carbon atom 
as a letter C then you must add all 
the H atoms too. If you don't want 
to draw all the Hs, don't write C 
for carbon. 



Be clear 

Try drawing some of the amino acids represented on p. 000 in a similar way, using the three guide- 
lines. The bond angles at tetrahedral carbon atoms are about 109°. Make them look about 109° pro- 
jected on to a plane! (120° is a good compromise, and it makes the drawings look neat.) 

Start with leucine — earlier we drew it as the structure to the right. Get a piece of paper and do it 
now; then see how your drawing compares with our suggestions. 



H. 
H 3 C 



CH 3 



H NH, 

/\ II 

H H U 



OH 



24 



2 ■ Organic structures 



It doesn't matter which way up you've drawn it, but your diagram should look something 
like one of these structures below. 



NH, 




leucine 



OH 




NH 



H0 2 C 




leucine 



HOOC 




leucine 

The guidelines we gave were only guidelines, not rules, and it certainly does not matter which way 
round you draw the molecule. The aim is to keep the functional groups clear, and let the skeleton 
fade into the background. That's why the last two structures are all right — the carbon atom shown as 
'C is part of a functional group (the carboxyl group) so it can stand out. 

Now turn back to p. 000 and try redrawing the some of the other eight structures there using the 
guidelines. Don't look at our suggestions below until you've done them! Then compare your draw- 
ings with our suggestions. 



H 2 N 



OH 





glycine 



NH 2 




OH 





alanine 




phenylalanine 




proline 



Not all chemists put circles round their 
plus and minus charges — it's a matter 
of personal choice. 



The wiggly line is a graphical way of 

indicating an incomplete structure: it 

shows where we have 

mentally 'snapped 

off' the C0 2 H group 

from the rest of the 

molecule. 



A 



r 





H,N 




tryptophan 



methionine 



lysine 



Remember that these are only suggestions, but we hope you'll agree that this style of diagram looks 
much less cluttered and makes the functional groups much clearer than the diagrams on p. 000. 
Moreover, they still bear significant resemblance to the 'real thing' — compare these crystal structures 
of lysine and tryptophan with the structures shown above, for example. 

Structural diagrams can be modified to suit the occasion 

You'll probably find that you want to draw the same molecule in different ways on different occa- 
sions to emphasize different points. Let's carry on using leucine as an example. We mentioned before 
that an amino acid can act as an acid or as a base. When it acts as an acid, a base (for example, 
hydroxide, OH~) removes H + from the carboxylic acid group in a reaction we can represent as 





H 2 



The product of this reaction has a negative charge on an oxygen atom. We have put it in a circle to 
make it clearer, and we suggest you do the same when you draw charges: +'s and -'s are easily 
mislaid. We shall discuss this type of reaction, the way in which reactions are drawn, and what the 
'curly arrows' in the diagram mean in Chapter 5. But for now, notice that we drew out the CO2H as 
the fragment left because we wanted to show how the O-H bond was broken when the base attacked. 
We modified our diagram to suit our own purposes. 



Drawing molecules 



25 



When leucine acts as a base, the amino (NH 2 ) group is involved. The nitrogen atom attaches itself 
to a proton, forming a new bond using its lone pair. 
We can represent this reaction as 
H 



N: 







H 



©T 



C0 2 H 




H,0 



C0 2 H 



Notice how we drew the lone pair at this time because we wanted to show how it was involved in 
the reaction. The oxygen atoms of the carboxylic acid groups also have lone pairs but we didn't draw 
them in because they weren't relevant to what we were talking about. Neither did we feel it was nec- 
essary to draw CO2H in full this time because none of the atoms or bonds in the carboxylic acid 
functional group was involved in the reaction. 

Structural diagrams can show three-dimensional information on a 
two-dimensional page 

Of course, all the structures we have been drawing only give an idea of the real structure of the 
molecules. For example, the carbon atom between the NH2 group and the CO2H group of 
leucine has a tetrahedral arrangement of atoms around it, a fact which we have so far completely 
ignored. 

We might want to emphasize this fact by drawing in the hydrogen atom we missed out at this 
point as in structure 1 (in the right-hand margin). 

We can then show that one of the groups attached to this carbon atom comes towards us, out of 
the plane of the paper, and the other one goes away from us, into the paper. There are several ways of 
doing this. In structure 2, the bold, wedged bond suggests a perspective view of a bond coming 
towards you, while the hashed bond suggests a bond fading away from you. The other two 'normal' 
bonds are in the plane of the paper. 

Alternatively we could miss out the hydrogen atom and draw something a bit neater though 
slightly less realistic as structure 3. 

We can assume the missing hydrogen atom is behind the plane of the paper, because that is where 
the 'missing' vertex of the tetrahedron of atoms attached to the carbon atom lies. These conventions 
allow us to give an idea of the three-dimensional shape (stereochemistry) of any organic molecule — 
you have already seen them in use in the diagram of the structure of palytoxin at the beginning of this 
chapter. 

• Reminder 

Organic structures should be drawn to be realistic, economical, clear. 

We gave three guidelines to help you achieve this when you draw structures: 

• Guideline 1: Draw chains of atoms as zig-zags 

• Guideline 2: Miss out the Hs attached to carbon atoms, along with the C-H 
bonds 

• Guideline 3: Miss out the capital Cs representing carbon atoms 

The guidelines we have given and conventions we have illustrated in this section have grown up 
over decades. They are used by organic chemists because they work! We guarantee to follow them for 
the rest of the book — try to follow them yourself whenever you draw an organic structure. Before 
you ever draw a capital C or a capital H again, ask yourself whether it's really necessary! 

Now that we have considered how to draw structures, we can return to some of the structural 
types that we find in organic molecules. Firstly, we'll talk about hydrocarbon frameworks, then 
about functional groups. 



A lone pair is a pair of electrons that is 
not involved in a chemical bond We 
shall discuss lone pairs in detail in 
Chapter 4. Again, don't worry about 
what the curly arrows in this diagram 
mean — wewillcoverthem in detail in 
Chapter 5. 




When you draw diagrams like 
these to indicate the three- 
dimensional shape of the 
molecule, try to keep the 
hydrocarbon framework in the 
plane of the paper and allow 
functional groups and other 
branches to project forwards out 
of the paper or backwards into it. 



We shall look in more detail at the 
shapes of molecules — their 
stereochemistry — in Chapter 16. 



26 



2 ■ Organic structures 



Hydrocarbon frameworks 



Carbon as an element is unique in the variety of structures it can form. It is unusual because it forms 
strong, stable bonds to the majority of elements in the periodic table, including itself. It is this ability 
to form bonds to itself that leads to the variety of organic structures that exist, and indeed to the pos- 
sibility of life existing at all. Carbon may make up only 0.2% of the earth's crust, but it certainly 
deserves a whole branch of chemistry all to itself. 

Chains 

The simplest class of hydrocarbon frameworks contains just chains of atoms. The fatty acids we met 
earlier have hydrocarbon frameworks made of zig-zag chains of atoms, for example. Polythene is a 
polymer whose hydrocarbon framework consists entirely of chains of carbon atoms. 



a section of the structure of polythene 



Notice we've drawn in four groups 
as CH 3 — we did this because we 
didn't want them to get 
overlooked in such a large 
structure. They are the only tiny 
branches off this long winding 
trunk. 



The names for shorter chains 
(which you must learn) exist for 
historical reasons; for chains of 5 
or more carbon atoms, the 
systematic names are based on 
Greek number names. 



At the other end of the spectrum of complexity is this antibiotic, extracted from a fungus in 1995 
and aptly named linearmycin as it has a long linear chain. The chain of this antibiotic is so long that 
we have to wrap it round two corners just to get it on the page. 

We haven't drawn H 2 N.. 
whether the CH 3 
groups and OH groups 
are in front of or 
behind the plane of the 
paper, because (at the 
time of writing this 
book) no one yet 
knows. The stereo- 
chemistry of linear- 
mycin is unknown. 

Names for carbon chains 

It is often convenient to refer to a chain of carbon atoms by a name indicating its length. You have 
probably met some of these names before in the names of the simplest organic molecules, the 
alkanes. There are also commonly used abbreviations for these names: these can be very useful in 
both writing about chemistry and in drawing chemical structures, as we shall see shortly. 




C0 2 H 



Names and abbreviations for carbon chains 






Number of carbon 


Name of 


Formula' 


Abbreviation 


Name of alkane 


atoms in chain 


group 






(= chain + H) 


1 


methyl 


-CH 3 


Me 


methane 


2 


ethyl 


-CH 2 CH 3 


Et 


ethane 


3 


propyl 


— CH2CH2CH3 


Pr 


propane 


4 


butyl 


-(CH 2 ) 3 CH 3 


Bu 


butane 


5 


pentyl 


-(CH 2 ) 4 CH 3 


_t 


pentane 


6 


hexyl 


-(CH 2 ) 5 CH 3 


_t 


hexane 


7 


heptyl 


-(CH 2 ) 6 CH 3 


_t 


heptane 


8 


octyl 


-(CH 2 ) 7 CH 3 


_f 


octane 


9 


nonyl 


-(CH 2 ) 8 CH 3 


_f 


nonane 


10 


decyl 


-(CH 2 )gCH 3 


_t 


decane 



This representation is not recommended. 
' Names for longer chains are not commonly abbreviated. 



Hydrocarbon frameworks 



27 



Organic elements 

You may notice that the abbreviations for the names of carbon chains look very much like the 
symbols for chemical elements: this is deliberate, and these symbols are sometimes called 'organic 
elements'. They can be used in chemical structures just like element symbols. It is often convenient 
to use the 'organic element' symbols for short carbon chains for tidiness. Here are some examples. 
Structure 1 to the right shows how we drew the structure of the amino acid methionine on p. 24. The 
stick representing the methyl group attached to the sulfur atom does, however, look a little odd. 
Most chemists would draw methionine as structure 2, with 'Me' representing the CH3 (methyl) 
group. Tetraethyllead used to be added to petrol to prevent engines 'knocking', until it was shown to 
be a health hazard. Its structure (as you might easily guess from the name) is shown as item 3. But it's 
much easier to write as PbEt 4 or Et 4 Pb 

Remember that these symbols (and names) can only be used for terminal chains of atoms. We 
couldn't abbreviate the structure of lysine from 





H,N 



lysine 
for example, because Bu represents 



to 




NOT CORRECT 



H H H H 



H 




H H H H 



/ 



and not 



fi 



H H H H 




•/ 



methionine 



-Pb- 



3 tetraethyllead 




H H H H 



Before leaving carbon chains, we must mention one other very useful organic element symbol, R. 
R in a structure can mean anything — it's a sort of wild card. For example, structure 4 would indicate 
any amino acid, where R = H is glycine, R = Me is alanine. . . As we've mentioned before, and you will 
see later, the reactivity of organic molecules is so dependent on their functional groups that the rest 
of the molecule can be irrelevant. In these cases, we can choose just to call it R. 

Carbon rings 

Rings of atoms are also common in organic structures. You may have heard the famous 
story of Auguste Kekule first realizing that benzene has a ring structure when he dreamed 
of snakes biting their own tails. You have met benzene rings in phenylalanine and aspirin. 
Paracetamol also has a structure based on a benzene ring. 




benzene 






aspirin 



phenylalanine paracetamol 

When a benzene ring is attached to a molecule by only one of its carbon 
atoms (as in phenylalanine, but not paracetamol or aspirin), we can call it a 
'phenyl' group and give it the organic element symbol Ph. 




the phenyl group, Ph 




is 

equivalent 

to 




Kek.j ^ s sn^ke dream inspired this fip..'^ 
that appear ad In a spoof edition of the 
German chemical Journal, Berfctoe der 
Devtscten Chemixten Gesetlsdmft 
inlBS6 



Benzene has a ring structure 

In 1865, August Kekule presented a paper at the Academie des 
Sciences in Paris suggesting a cyclic structure for benzene, the 
inspiration for which he ascribed to a dream. However, was 
Kekule the first to suggest that benzene was cyclic? Some 
believe not, and credit an Austrian schoolteacher, Josef 
Loschmidt with the first depiction of cyclic benzene structures. 
In 1861, 4 years before Kekule's 
dream, Loschmidt published a book in 
which he represented benzene as a set 
of rings. It is not certain whether 
Loschmidt or Kekule — or even a Scot 
named Archibald Couper — got it right 
first. 




28 



2 ■ Organic structures 



Any compound containing a benzene ring, or a related (Chapter 7) ring system is known as 'aro- 
matic', and another useful organic element symbol related to Ph is Ar (for 'aryl'). While Ph always 
means CsHs, Ar can mean any substituted phenyl ring, in other words, phenyl with any number of 
the hydrogen atoms replaced by other groups. 

For example, while PhOH always means phenol, ArOH could mean phenol, 2,4,6-trichlorophe- 
nol (the antiseptic TCP), paracetamol or aspirin (among many other substituted phenols). Like R, 



Of course, Ar= argon too, but so 
few argon compounds exist that 
there is never any confusion. 



PhOH 




phenol 



the 'wild card' alkyl group, Ar is a 'wild card' aryl 
group. 

The compound known as muscone has only rel- 
atively recently been made in the lab. It is the pun- 
gent aroma that makes up the base-note of musk 
fragrances. Before chemists had determined its 
structure and devised a laboratory synthesis the 
only source of musk was the musk deer, now rare 
for this very reason. Muscone's skeleton is a 13- 
membered ring of carbon atoms. 

The steroid hormones have several (usually four) rings fused together. These are testosterone and 
oestradiol, the important human male and female sex hormones. 




phenol 



2,4,6-trichlorophenol 




testosterone 



oestradiol 



Some ring structures are much more complicated. The potent poison strychnine is a tangle of 
interconnecting rings. 

Buckminsterfullerene 

Buckminsterfullerene is named 
after the American inventor and 
architect Richard Buckminster 
Fuller, who designed the structures 
known as 'geodesic domes'. 





Buckminsterfullerene 




One of the most elegant ring structures is shown above and is known as Buckminsterfullerene. It 
consists solely of 60 carbon atoms in rings that curve back on themselves to form a football-shaped 
cage. 

Count the number of bonds at any junction and you will see they add up to four so no hydrogens 
need be added. This compound is Cgg. Note that you can't see all the atoms as some are behind the 
sphere. 

Rings of carbon atoms are given names starting with 'cyclo', followed by the name for the carbon 
chain with the same number of carbon atoms. 



Hydrocarbon frameworks 



29 



H °Y 




To the right, structure 1 shows chrysanthemic acid, part of the naturally occurring pesticides 
called pyrethrins (an example appears in Chapter 1), which contains a cyclopropane ring. Propane 
has three carbon atoms. Cyclopropane is a three-membered ring. Grandisol (structure 2), an insect 
pheromone used by male boll weevils to attract females, has a structure based on a cyclobutane ring. 
Butane has four carbon atoms. Cyclobutane is a four-membered ring. Cyclamate (structure 3), for- 
merly used as an artificial sweetener, contains a cyclohexane ring. Hexane has six carbon atoms. 
Cyclohexane is a six-membered ring. 



1 chrysanthemic acid 



Branches 

Hydrocarbon frameworks rarely consist of single rings or chains, but are often branched. Rings, 
chains, and branches are all combined in structures like that of the marine toxin palytoxin that we 
met at the beginning of the chapter, polystyrene, a polymer made of six-membered rings dangling 
from linear carbon chains, or of (3-carotene, the compound that makes carrots orange. 




part of the structure of polystyrene 





grandisol 




^S0 3 H 



cyclamate 



p-carotene 



Just like some short straight carbon chains, some short branched carbon chains are given names and 
organic element symbols. The most common is the isopropyl group. Lithium diisopropylamide 
(also called LDA) is a strong base commonly used in organic synthesis. 
Li 



the isopropyl group 
/-Pr 




is equivalent to LiN/'-Pr 2 



lithium diisopropylamide (LDA) 

Iproniazid is an antidepressant drug with i-Pr in both structure and name. 

Notice how the 'propyl' part of 'isopropyl' still indicates three 
carbon atoms; they are just joined together in a different way — in 
other words, as an isomer of the straight chain propyl group. 
Sometimes, to avoid confusion, the straight chain alkyl groups are 
called '«-alkyl' (for example, n-Pr, n-Bu) — n for 'normal' — to 
distinguish them from their branched counterparts. 




i-PrHN 

is equivalent to 



proniazid 




Isomers are molecules with the same kinds and numbers of atoms joined up in 
different ways n-propanol, n-PrOH, andisopropanol, i-PrOH, are isomeric 
alcohols. Isomers need not have the same functional groups, these compounds are 
all isomers of C 4 H 8 0. 

OH 

CHO 





'Isopropyl' may be abbreviated to /-Pr, 
/Pr, or Pr'. We will use the first in this 
book, but you may see the others used 
elsewhere. 



30 



2 ■ Organic structures 



The isobutyl (i-Bu) group is a CH 2 group joined to an z'-Pr group. It is z'-PrCH 2 - 
Two isobutyl groups are present in the reducing agent diisobutyl aluminium hydride (DIBAL). 
H 

Al 




the isobutyl group 
i-Bu 





Notice how the invented name 
ibuprofen is a medley of 'ibu' (from i-Bu 
for isobutyl) + 'pro' (for propyl, the 
three-carbon unit shown in gold) + 'fen' 
(forthe phenyl ring). We will talk about 
the way in which compounds are 
named later in this chapter. 



/- 





the sec-butyl group 
s-Bu 



A 




the tert-butyl group 
f-Bu 



diisobutyl aluminium hydride (DIBAL) 
is equivalent to HAIi-Bu 2 

The painkiller ibuprofen (marketed as Nurofen ) contains an isobutyl group. 



C0 2 H 



Ibuprofen 

There are two more isomers of the butyl group, both of which have common names and abbrevi- 
ations. The sec-butyl group (s-butyl or s-Bu) has a methyl and an ethyl group joined to the same car- 
bon atom. It appears in an organolithium compound, sec-butyl lithium, used to introduce lithium 
atoms into organic molecules. 

Li 

is equivalent to s-BuLi 

The ferf-butyl group (f-butyl or f-Bu) group has three methyl groups joined to the same carbon 
atom. Two t-Bu groups are found in BHT ('butylated hydroxy toluene'), an antioxidant added to 
some processed foods. 

OH i OH 





t-Bu 



BHT 




t-Bu 



i Primary, secondary, and tertiary 

The prefixes sec and tertare really short for secondary and tertiary, terms that refer 
to the carbon atom that attaches these groups to the rest of the molecular 
structure. 



methyl 
(no attached C) 



primary 
(1 attached C) 



fl 



I 



secondary 
(2 attached C) 



Me — OH 

methanol 



butan-1-ol 
n-butanol 



^N)H 

butan-2-ol 
sec-butanol 



tertiary 
(3 attached C) 



OH 




quaternary 
(4 attached C) 



OH 




2-methypropan-2-ol 2,2,-dimethylpropan-l-ol 
tert-butanol 



A primary carbon atom is attached to only one other C atom, a secondary to two 
other C atoms, and so on. This means there are five types of carbon atom. 

These names for bits of hydrocarbon framework are more than just useful ways of 
writing or talking about chemistry. They tell us something fundamental about the 
molecule and we shall use them when we describe reactions. 



Functional groups 



31 



This quick architectural tour of some of the molecular edifices built by nature and by man serves 
just as an introduction to some of the hydrocarbon frameworks you will meet in the rest of this chap- 
ter and of this book. Yet, fortunately for us, however complicated the hydrocarbon framework might 
be, it serves only as a support for the functional groups. And, by and large, a functional group in one 
molecule behaves in much the same way as it does in another molecule. What we now need to do, 
and we start in the next section, is to introduce you to some functional groups, and to explain why it 
is that their attributes are the key to understanding organic chemistry. 



Functional groups 



If you can take ethane gas (CH3CH3, or EtH, or even ^^ , though a single line like this doesn't 
look much like a chemical structure) and bubble it through acids, bases, oxidizing agents, reducing 
agents — in fact almost any chemical you can think of — it will remain unchanged. Just about the only 
thing you can do with it is burn it. Yet ethanol (CH3CH2OH, or \ 



only burns, it reacts with acids, bases, and oxidizing agents. 

The difference between ethanol and ethane is the functional 
group — the OH or hydroxyl group. We know that these chemical 
properties (being able to react with acids, bases, and oxidizing 
agents) are properties of the hydroxyl group and not just of 
ethanol because other compounds containing OH groups (in 
other words, other alcohols) have similar properties, whatever 
their hydrocarbon frameworks. 

Your understanding of functional groups will be the key to your 
understanding of organic chemistry. We shall therefore now go on to 
meet some of the most important functional groups. We won't say 
much about the properties of each group; that will come in Chapter 5 
and later. Your task at this stage is to learn to recognize them when 
they appear in structures, so make sure you learn their names. The 
classes of compound associated with some functional groups also 
have names: for example, compounds containing the hydroxyl group 
are known as alcohols. Learn these names too as they are more 
important than the systematic names of individual compounds. 
We've told you a few snippets of information about each group to 
help you to get to know something of the group's character. 



, or preferably EtOH) not 
Ethanol 

The reaction of ethanol with oxidizing agents makes vinegar from wine and 
sober people from drunk ones. In both cases, the oxidizing agent is oxygen 
from the air, catalysed by an enzyme in a living system. The oxidation of 
ethanol by microorganisms that grow in wine left open to the air leads to 
acetic acid (ethanoic acid) while the oxidation of ethanol by the liver gives 
acetaldehyde (ethanal). 

^ \ ^OH w 



micro- 
organism 



acetaldehyde 



ethanol 



OH 

acetic acid 



Human metabolism and oxidation 

The human metabolism makes use of the oxidation of alcohols to render 
harmless other toxic compounds containing the OH group. For example, lactic 
acid, produced in muscles during intense activity, is oxidized by an enzyme called 
lactate dehydrogenase to the metabolically useful compound pyruvic acid. 

OH 

2 



v C0 2 H 

lactic acid 



lactate dehydrogenase 



X0 2 H 

pyruvic acid 



Alkanes contain no functional groups 

The alkanes are the simplest class of organic molecules because they contain no functional groups. 
They are extremely unreactive, and therefore rather boring as far as the organic chemist is con- 
cerned. However, their unreactivity can be a bonus, and alkanes such as pentane and hexane are 
often used as solvents, especially for purification of organic compounds. Just about the only thing 
alkanes will do is burn — methane, propane, and butane are all used as domestic fuels, and petrol is a 
mixture of alkanes containing largely isooctane. 




pentane 



hexane 



isooctane 



Alkenes (sometimes called olefins) contain C=C double bonds 

It may seem strange to classify a type of bond as a functional group, but you will see later that C=C 
double bonds impart reactivity to an organic molecule just as functional groups consisting of, say, 
oxygen or nitrogen atoms do. Some of the compounds produced by plants and used by perfumers 
are alkenes (see Chapter 1). For example, pinene has a smell evocative of pine forests, while limonene 
smells of citrus fruits. 




a-pmene 



^^ 



32 



2 ■ Organic structures 



You've already met the orange pigment (3-carotene. Eleven C=C double bonds make up most of 
its structure. Coloured organic compounds often contain chains of C=C double bonds like this. In 
Chapter 7 you will find out why this is so. 




P-carotene 



► Saturated and 

unsaturated carbon atoms 

In an alkane, each carbon atom is 
joined to four other atoms (C or 
H). It has no potential for forming 
more bonds and is therefore 
saturated. In alkenes, the carbon 
atoms making up the C=C double 
bond are attached to only three 
atoms each. They still have the 
potential to bond with one more 
atom, and are therefore 
unsaturated. In general, carbon 
atoms attached to four other 
atoms are saturated; those 
attached to three, two, or one are 
unsaturated. 



Alkynes contain C^C triple bonds 

Just like C=C double bonds, C=C triple bonds have a 
special type of reactivity associated with them, so it's 
useful to call a C=C triple bond a functional group. 
Alkynes are linear so we draw them with four carbon 
atoms in a straight line. Alkynes are not as widespread 
in nature as alkenes, but one fascinating class of com- 
pounds containing C=C triple bonds is a group of 
antitumour agents discovered during the 1980s. 
Calicheamicin is a member of this group. The high 
reactivity of this combination of functional groups 
enables calicheamicin to attack DNA and prevent can- 
cer cells from proliferating. For the first time we have 
drawn a molecule in three dimensions, with two 
bonds crossing one another — can you see the shape? 



MeO 







calicheamicin 
(R = a string of sugar molecules) 



Remember that R can mean any alkyl 
group. 



If we want a structure to contain more 
than one 'R', we give the R's numbers 
and call them R 1 , R 2 .. . Thus R^O-R 2 
means an ether with two different 
unspecified alkyl groups. (NotR-L, R2--, 
which would mean lxR, 2xR...) 



diethyl ether 

"ether" THF 



HO 



Alcohols (R-OH) contain a hydroxyl (OH) group 

We've already talked about the hydroxyl group in ethanol and other alcohols. Carbohydrates are 
peppered with hydroxyl groups; sucrose has eight of them for example (a more three-dimensional 
picture of the sucrose molecule appears in Chapter 1). 
Molecules containing hydroxyl groups are often 
soluble in water, and living things often attach sugar 
groups, containing hydroxyl groups, to otherwise 
insoluble organic compounds to keep them in solu- 
tion in the cell. Calicheamicin, a molecule we have just 
mentioned, contains a string of sugars for just this rea- 
son. The liver carries out its task of detoxifying 
unwanted organic compounds by repeatedly hydroxy - 
lating them until they are water-soluble, and they are 
then excreted in the bile or urine. 



HO^Y"""'OH 
OH 




Another common laboratory solvent is 
called 'petroleum ether'. Don't confuse 
this with diethyl ether! Petroleum ether 
is in fact not an ether, but a mixture of 
alkanes. 'Ether', according to the 
Oxford English Dictionary, means 'clear 
sky, upper region beyond the clouds', 
and hence used to be used for anything 
light, airy, and volatile. 



Ethers (R -O-R ) contain an alkoxy group (-OR) 

The name ether refers to any compound that has two alkyl groups linked through an oxygen atom. 
'Ether' is also used as an everyday name for diethyl ether, Et 2 0. You might compare this use of the 
word 'ether' with the common use of the word 'alcohol' to mean ethanol. Diethyl ether is a highly 
flammable solvent that boils at only 35 °C. It used to be used as an anaesthetic. Tetrahydrofuran 
(THF) is another commonly used solvent and is a cyclic ether. 

Brevetoxin B is a fascinating naturally occurring compound that was synthesized in the labora- 
tory in 1995. It is packed with ether functional groups in ring sizes from 6 to 8. 



Functional groups 



33 



Brevetoxin B 

Brevetoxin B is one of a family of polyethers found in a sea 
creature (a dinoflagellate Gymnodinium breve, hence the 
name) which sometimes multiplies at an amazing rate and 
creates 'red tides' around the coasts of the Gulf of Mexico. 
Fish die in shoals and so do people if they eat the shellfish 
that have eaten the red tide. The brevetoxins are the killers 
The many ether oxygen atoms interfere with sodium ion 
(Na + ) metabolism. 




Amines (R-NH2) contain the amino (NH2) group 

We met the amino group when we were discussing the amino acids: we mentioned that it was this 
group that gave these compounds their basic properties. Amines often have powerful fishy smells: the 
smell of putrescine is particularly foul. It is formed as meat decays. Many neurologically active com- 
pounds are also amines: amphetamine is a notorious stimulant. 

Nitro compounds (R-NO2) contain the nitro group (NO2) 

The nitro group (NO2) is often incorrectly drawn with five bonds to nitrogen which you will see in 
Chapter 4, is impossible. Make sure you draw it correctly when you need to draw it out in detail. If 
you write just NO2 you are all right! 

Several nitro groups in one molecule can make it quite unstable and even explosive. Three nitro 
groups give the most famous explosive of all, TNT (trinitrotoluene), its kick. 



H 2 N 




amphetamine 



the nitro 
group 



2 N 




TNT 




nitrogen cannot 
have five bonds! 



■r-'^o 

incorrect structure 
for the nitro group 



nitrazepam 



However, functional groups refuse to be stereotyped. Nitrazepam also contains a nitro group, but 
this compound is marketed as Mogadon , the sleeping pill. 

Alkyl halides (fluorides R-F, chlorides R-Cl, bromides R-Br, or iodides R-I) 
contain the fluoro, chloro, bromo, or iodo groups 

These three functional groups have similar properties — though alkyl iodides are the most reactive 
and alkyl fluorides the least. PVC (polyvinyl chloride) is one of the most widely used polymers — it 
has a chloro group on every other carbon atom along a linear hydrocarbon framework. Methyl 
iodide (Mel), on the other hand, is a dangerous carcinogen, since it reacts with DNA and can cause 
mutations in the genetic code. 




CI CI CI CI 

a section of the structure of PVC 



These compounds are also known as 
haloalkanes (fluoroalkanes, 
chloroalkanes, bromoalkanes or 
iodoalkanes). 



Because alkyl halides have 
similar properties, chemists use 
yet another 'wild card' organic 
element, X, as a convenient 
substitute for CI, Br, or I 
(sometimes F). So R-X is any alkyl 
halide. 



34 



2 ■ Organic structures 



-CHO represents: 



1 •} 

Aldehydes (R-CHO) and ketones (R -CO-R ) contain the carbonyl group C=0 

Aldehydes can be formed by oxidizing alcohols — in fact the liver detoxifies ethanol in the bloodstream 
by oxidizing it first to acetaldehyde (ethanal, CH3CHO). Acetaldehyde in the blood is the cause of hang- 
overs. Aldehydes often have pleasant smells — 2-methylundecanal is a key component of the fragrance of 
Chanel No 5™, and 'raspberry ketone' is the major component of the flavour and smell of raspberries. 

Q 



X 






When we write aldehydes as R-CHO, 
we have no choice but to write in the C 
and H (because they're part of the 
functional group) — one important 
instance where you should ignore 
Guideline 3 for drawing structures. 
Another point: always write R-CHO and 
never R-COH, which looks too much 
like an alcohol. 





2-methylundecanal 



"raspberry ketone" 



Carboxylic acids (R-CO2H) contain the carboxyl group CO2H 

As their name implies, compounds containing the carboxylic acid (CO2H) group can react with bases, 
losing a proton to form carboxylate salts. Edible carboxylic acids have sharp flavours and several are 
found in fruits — citric, malic, and tartaric acids are found in lemons, apples, and grapes, respectively. 



OH 



ho 2 <t yr ^co 2 h 

HO C0 2 H 

citric acid 



H0 2 C 



C0 2 H 



H0 2 C 



OH 

malic acid 




C0 2 H 



OH 

tartaric acid 



The terms 'saturated fats' and 
'unsaturated fats' are familiar — they 
refer to whether the R groups are 
saturated (no C=C double bonds) or 
unsaturated (contains C=C double 
bonds) — see the box on p. 000. Fats 
containing R groups with several 
double bonds (for example, those that 
are esters formed from linoleic acid, 
which we met at the beginning of this 
chapter) are known as 
'polyunsaturated'. 



Esters (R -CO2R ) contain a carboxyl group with an extra alkyl group (CO2R) 

Fats are esters; in fact they contain three ester groups. 
They are formed in the body by condensing glycerol, a 
compound with three hydroxyl groups, with three 
fatty acid molecules. 

Other, more volatile esters, have pleasant, fruity 
smells and flavours. These three are components of 
the flavours of bananas, rum, and apples: 








isopentyl acetate 
(bananas) 



isobutyl propionate 
(rum) 



a fat molecule 
a long alkyl chain) 




isopentyl valerate 
(apples) 



Amides (R-CONH 2 , R^CONHR 2 , or R^ONR^ 3 ) 

Proteins are amides: they are formed when the carboxylic acid group of one amino acid condenses 
with the amino group of another to form an amide linkage (also known as a peptide bond). One pro- 
tein molecule can contain hundreds of amide bonds. Aspartame, the artificial sweetener marketed as 
NutraSweet , on the other hand contains just two amino acids, aspartic acid and phenylalanine, 
joined through one amide bond. Paracetamol is also an amide. 



H0 2 C 



OMe 





aspartame 



paracetamol 



Carbon atoms carrying functional groups can be classified by oxidation level 



35 




Nitriles or cyanides (R-CN) contain the cyano group -C=N 

Nitrile groups can be introduced into molecules by reacting potassium cyanide with alkyl halides. 
The organic nitrile group has quite different properties associated with lethal inorganic cyanide: 
Laetrile, for example, is extracted from apricot kernels, and was once developed as an anticancer 
drug. It was later proposed that the name be spelt 'liar-trial' since the results of the clinical trials on 
laetrile turned out to have been falsified! 

Acyl chlorides (acid chlorides) (R-COC1) 

Acyl chlorides are reactive compounds used to make esters and amides. They are derivatives of car- 
boxylic acids with the -OH replaced by —CI, and are too reactive to be found in nature. 

Acetals 

Acetals are compounds with two single bonded oxygen atoms attached to the same carbon atom. 
Many sugars are acetals, as is laetrile which you have just met. 



HO 



A, 



acetyl chloride 



RO x 



an acetal 



OR HO 



/v^y 0. 



HOT^r^ '"OH 

1 OH OH 

OH 

sucrose 




Carbon atoms carrying functional groups can be classified by 
oxidation level 

All functional groups are different, but some are more different than others. For example, the struc- 
tures of a carboxylic acid, an ester, and an amide are all very similar: in each case the carbon atom car- 
rying the functional group is bonded to two heteroatoms, one of the bonds being a double bond. You 
will see in Chapter 12 that this similarity in structure is mirrored in the reactions of these three types of 
compounds, and in the ways in which they can be interconverted. Carboxylic acids, esters, and amides 
can be changed one into another by reaction with simple reagents such as water, alcohols, or amines 
plus appropriate catalysts. To change them into aldehydes or alcohols requires a different type or 
reagent, a reducing agent (a reagent which adds hydrogen atoms). We say that the carbon atoms car- 
rying functional groups that can be interconverted without the need for reducing agents (or oxidizing 
agents) have the same oxidation level — in this case, we call it the 'carboxylic acid oxidation level'. 



The carboxylic acid 
oxidation level 



o u U 

XXX 

^^OH FT^OR' R ^NH 2 



^ 



V 0H 

carboxylic acids 



u 
*X C| 



esters 



amides 



nitriles acyl chlorides 



In fact, amides can quite easily be converted into nitriles just by dehydration (removal of water), so 
we must give nitrile carbon atoms the same oxidation level as carboxylic acids, esters, and amides. 
Maybe you're beginning to see the structural similarity between these four functional groups that 
you could have used to assign their oxidation level? In all four cases, the carbon atom has three bonds 
to heteroatoms, and only one to C or H. It doesn't matter how many heteroatoms there are, just how 
many bonds to them. Having noticed this, we can also assign both carbon atoms in 'CFC-113', one of 
the environmentally unfriendly aerosol propellants/refrigerants that have caused damage to the 
earth's ozone layer, to the carboxylic acid oxidation level. 

Aldehydes and ketones contain a carbon atom with two bonds to heteroatoms; they are at the 
'aldehyde oxidation level'. The common laboratory solvent dichloromethane also has two bonds to 
heteroatoms, so it too contains a carbon atom at the aldehyde oxidation level, as do acetals. 



► A heteroatom is an atom 
that is not C or H 

You've seen that a functional 
group is essentially any deviation 
from an alkane structure, either 
because the molecule has fewer 
hydrogen atoms than an alkane 
(alkenes, alkynes) or because it 
contains a collection of atoms 
that are not C and not H . There is 
a useful term for these 'different' 
atoms: heteroatoms. A 
heteroatom is any atom in an 
organic molecule other than Cor 
H. 



Don't confuse oxidation level with 
oxidation state. In all of these 
compounds, carbon is in 
oxidation state +4. 



F — C— C— CI 

/ \ 
CI CI 



"CFC-113" 



36 



2 ■ Organic structures 



The aldehyde 
oxidation level 






R "H 

aldehydes 



R "R' 

ketones 



RO OR 

r^r- 



acetals 



C '\/ Cl 



dichloromethane 



Alcohols, ethers, and alkyl halides have a carbon atom with only one single bond to a heteroatom. We 
assign these the 'alcohol oxidation level', and they are all easily made from alcohols without oxida- 
tion or reduction. 



The alcohol 
oxidation level 



R "OH 

alcohols 



R "OR' 

ethers 



"CI R Br 

alkyl halides 



Lastly, we must include simple alkanes, which have no bonds to heteroatoms, as an 'alkane oxidation 
level'. 



• The alkane 
oxidation level 


H H 
H /C ^H 




methane 



The small class of compounds that have a carbon atom with four bonds to heteroatoms is related 
to CO2 and best described as at the carbon dioxide oxidation level. 



X 

cr xi 

"CFC-12" 

one of the refrigerants/ 

aerosol propellants which 

has caused damage to the 

earth's ozone layer 



• The carbon dioxide 




oxidation level 


CI CI 


II 

EtO^ ^OEt 
0=C=0 

diethyl carbonate 


v 

Cl^ ^Cl 

carbon tetrachloride 


useful reagent for 


formerly used as a 


carbon dioxide adding ester groups 


dry cleaning fluid 



• Summary: Important functional groups and oxidation levels 


Zero bonds 


One bond 


Two bonds 


Three bonds 


Four bonds 


to heteroatoms 


to heteroatom 


to heteroatoms 


to heteroatoms 


to heteroatoms 


Alkane 


Alcohol 


Aldehyde 


Carboxylic acid 


Carbon dioxide 


oxidation level 


oxidation level 


oxidation level 


oxidation level 


oxidation level 


R 2 R 3 

K 

R 1 ^R 4 


R^^^OH alcohols 
r"^OR' ethers 




aldehydes 
R^H 




carboxylic 
JL acids 
R N)H 


0=C=0 

carbon dioxide 


alkanes 














R^^NH 2 amines 


ketones 
R^R' 


esters 
R^^OR' 




II 

EtCT ^OEt 






RO OR 





diethyl carbonate 




R^^CI 


V acetals 
R'/%' 


11 amides 
R''^NH 2 






/\ alkyl 






P F 




R Br halides 






v 




R^^l 




& nitriles 


cr ^ci 

cFc-12 




R^^^> alkenes 


R ^^B alkynes 




acyl 
Jk.^ chlorides 







Systematic nomenclature 37 



Alkenes and alkynes obviously don't fit easily into these categories as they have no bonds to het- 
eroatoms. Alkenes can be made from alcohols by dehydration without any oxidation or reduction so 
it seems sensible to put them in the alcohol column. Similarly, alkynes and aldehydes are related by 
hydration/dehydration without oxidation or reduction. 



Naming compounds 



So far, we have talked a lot about compounds by name. Many of the names we've used (palytoxin, 
muscone, brevetoxin...) are simple names given to complicated molecules without regard for the 
actual structure or function of the molecule — these three names, for example, are all derived from 
the name of the organism from which the compound was first extracted. They are known as trivial 
names, not because they are unimportant, but because they are used in everyday scientific conversa- 
tion. 

Names like this are fine for familiar compounds that are widely used and referred to by chemists, 
biologists, doctors, nurses, perfumers alike. But there are over 16 million known organic com- 
pounds. They can't all have simple names, and no one would remember them if they did. For this 
reason, the IUPAC (International Union of Pure and Applied Chemistry) have developed systemat- 
ic nomenclature, a set of rules that allows any compound to be given a unique name that can be 
deduced directly from its chemical structure. Conversely, a chemical structure can be deduced from 
its systematic name. 

The problem with systematic names is that they tend to be grotesquely unpronounceable for any- 
thing but the most simple molecules. In everyday speech and writing, chemists therefore do tend to 
disregard them, and use a mixture of systematic and trivial names. Nonetheless, it's important to 
know how the rules work. We shall look next at systematic nomenclature, before going on to look at 
the real language of chemistry. 



Systematic nomenclature 



There isn't space here to explain all the rules for giving systematic names for compounds — they fill 
several desperately dull volumes, and there's no point knowing them anyway since computers will do 
the naming for you. What we will do is to explain the principles underlying systematic nomencla- 
ture. You should understand these principles, because they provide the basis for the names used by 
chemists for the vast majority of compounds that do not have their own trivial names. 

Systematic names can be divided into three parts: one describes the hydrocarbon framework; one 
describes the functional groups; and one indicates where the functional groups are attached to the 
skeleton. 

You have already met the names for some simple fragments of hydrocarbon framework (methyl, 
ethyl, propyl...). Adding a hydrogen atom to these alkyl fragments and changing -yl to -ane makes 
the alkanes and their names. You should hardly need reminding of their structures: 



Names for the hydrocarbon framework 



one carbon methane CH 4 



two carbons ethane CH 3 — CH 3 



three carbons propane CH 3 . ^CH 3 cyclopropane 



A 



38 



2 ■ Organic structures 



Names for the hydrocarbon framework (continued) 



cyclobutane 



□ 



five carbons 



pentane 



cyclopentane 



six carbons 



eight carbons 



hexane 



seven carbons heptane 



octane 



cyclohexane 



cycloheptane 



cyclo-octane 




nine carbons 



cyclononane 




ten carbons 



decane 



cyclodecane 




The name of a functional group can be added to the name of a hydrocarbon framework either as a 
suffix or as a prefix. Some examples follow. It is important to count all of the carbon atoms in the 
chain, even if one of them is part of a functional group: so pentanenitrile is actually BuCN. 



CH3OH 



methanol 



x 



ethanal 




cyclohexanone 



butanoic acid 



OH 



pentanenitrile 




heptanoyl chloride 



HC^CH 



ethyne 



ethoxyethane 



CH3NO2 



nitromethane 



^^ 



propene 



iodobenzene 




Compounds with functional groups attached to a benzene ring are named in a similar way. 

Numbers are used to locate functional groups 

Sometimes a number can be included in the name to indicate which carbon atom the functional 
group is attached to. None of the above list needed a number — check that you can see why not for 
each one. When numbers are used, the carbon atoms are counted from one end. In most cases, either 
of two numbers could be used (depending on which end you count from); the one chosen is always 
the lower of the two. Again, some examples will illustrate this point. Notice again that some func- 
tional groups are named by prefixes, some by suffixes, and that the number always goes directly 
before the functional group name. 



Systematic nomenclature 



39 



NH 2 

2' 



propan-1-ol 
OH 
2 J 



2-aminobutane 
NH 2 



pentan-2-one 
2 „ 



NOT 
CORRECT 


propan-2-ol (not 3-aminobutane) pentan-3-one 

Here are some examples of compounds with more than one functional group. 
NH 2 H 2 N. 



but-2-ene 



^C0 2 H 

2-aminobutanoic acid 



1,6-diaminohexane 



H0 2 C 

hexanedioic acid 



„C0 2 H 



CBr 4 

tetrabromomethane 



CI 



Again, the numbers indicate how far the functional groups are from the end of the carbon chain. 
Counting must always be from the same end for each functional group. Notice how we use di-, tri-, 
tetra- if there are more than one of the same functional group. 

With cyclic compounds, there isn't an end to the chain, but we can use numbers to show the dis- 
tance between the two groups — start from the carbon atom carrying one of the functional groups, 
then count round. 




OH 



2 N 



2-aminocyclohexanol 




2,4,6-trinitrobenzoic acid 



These rules work for hydrocarbon frameworks that are chains or rings, but many skeletons are 
branched. We can name these by treating the branch as though it were a functional group: 




2-methylbutane 




HO 



1-butylcyclopropanol 



1,3.5-trimethyl benzene 



Ortho, meta, and para 

With substituted benzene rings, an alternative way of identifying the positions of the substituents is to 
use the terms ortho, meta, and para. Ortho compounds are 1,2-disubstituted, meta compounds are 
1,3-disubstituted, and para compounds are 1,4-disubstituted. Some examples should make this clear. 




C0 2 H 




H 2 N 




OH 



1,2-dichlorobenzene 
or orthodichlorobenzene 
or o-dichlorobenzene 



3-chlorobenzoic acid 

or meta-chlorobenzoic acid 

or m-chlorobenzoic acid 



4-aminophenol 

or para-aminophenol 

or p-aminophenol 



Me' 



~*C\ 



CI 



1,1,1-trichloroethane 



ortho, meta, and para are often 
abbreviated to o, m, and p. 



40 



2 ■ Organic structures 



The terms ortho, meta, and para are used by chemists because they're easier to remember than 
numbers, and the words carry with them chemical meaning. 'Ortho' shows that two groups are next 
to each other on the ring even though the atoms may not happen to be numbered 1 and 2. They are 
one example of the way in which chemists don't always use systematic nomenclature but revert to 
more convenient 'trivial' terms. We consider trivial names in the next section. 



Beware! Ortho, meta, and para 
are used in chemistry to mean 
other things too: you may come 
across orthophosphoric acid, 
metastable states, and 
paraformaldehyde — these have 
nothing to do with the substitution 
patterns of benzene rings. 






Me'' ^OH 



We haven't asked you to 
remember any trivial names of 
molecules yet. But these 10 
compounds are so important, you 
must be able to rememberthem. 
Learn them now. 



What do chemists really call compounds? 

The point of naming a compound is to be able to communicate with other chemists. Most chemists 
are happiest communicating chemistry by means of structural diagrams, and structural drawings are 
far more important than any sort of chemical nomenclature. That's why we explained in detail how 
to draw structures, but only gave an outline of how to name compounds. Good diagrams are easy to 
understand, quick to draw, and difficult to misinterpret. 

# Always give a diagram alongside a name unless it really is something very simple, 
such as ethanol. 

But we do need to be able to communicate by speech and by writing as well. In principle we could 
do this by using systematic names. In practice, though, the full systematic names of anything but the 
simplest molecules are far too clumsy for use in everyday chemical speech. There are several alterna- 
tives, mostly based on a mixture of trivial and systematic names. 

Names for well known and widely used simple compounds 

A few simple compounds are called by trivial names not because the systematic names are complicat- 
ed, but just out of habit. We know them so well that we use their familiar names. 

You may have met this compound before (left), and perhaps called it ethanoic acid, its systematic 
name. But in a chemical laboratory, everyone would refer to this acid as acetic acid, its trivial name. 
The same is true for all these common substances. 

o P 

/\ M _ M »A» «^ nH Me-^OH EtO^ 



Me Me 

acetone 



Et 2 

ether, 
or diethyl ether 



Me'' "H 

acetaldehyde 



H ^OH 

formic acid 



Me'' ^OH 

acetic acid 



EtO' Me 

ethyl acetate 






benzene 



toluene 



phenol 




Trivial names like this are often long-lasting, well understood historical names that are less easy to 
confuse than their systematic counterparts. 'Acetaldehyde' is easier to distinguish from 'ethanol' 
than is 'ethanal'. 

Trivial names also extend to fragments of structures containing functional groups. Acetone, 
acetaldehyde, and acetic acid all contain the acetyl group (MeCO-, ethanoyl) abbreviated Ac and 
chemists often use this 'organic element' in writing AcOH for acetic acid or EtOAc for ethyl acetate. 

Chemists use special names for four fragments because they have mechanistic as well as structural 
significance. These are vinyl and allyl; phenyl and benzyl. 



<^ < 



°< 




A 




the vinyl group the allyl group 



the phenyl group: Ph the benzyl group: Bn 



What do chemists really call compounds? 



41 



Giving the vinyl group a name allows chemists to use simple trivial names for compounds like 
vinyl chloride, the material that polymerizes to give PVC (polyvinyl chloride) but the importance of 
the name lies more in the difference in reactivity (Chapter 17) between vinyl and allyl groups. 




CI 

vinyl chloride 



CI CI CI CI CI CI 

a section of the structure of PVC - Poly Vinyl Chloride 



The allyl group gets its name from garlic (Allium sp.), because it makes up part of the structure of 
the compounds responsible for the taste and smell of garlic. 

Allyl and vinyl are different in that the vinyl group is attached directly to a double bonded C=C 
carbon atom, while the allyl group is attached to a carbon atom adjacent to the C=C double bond. 
The difference is extremely important chemically: allyl compounds are typically quite reactive, while 
vinyl compounds are fairly unreactive. 

For some reason, the allyl and vinyl groups have never acquired organic element symbols, but the 
benzyl group has and is called Bn. It is again important not to confuse the benzyl group with the 
phenyl group: the phenyl group is joined through a carbon atom in the ring, while the benzyl group 
is joined through a carbon atom attached to the ring. Phenyl compounds are typically unreactive but 
benzyl compounds are often reactive. Phenyl is like vinyl and benzyl is like allyl. 





X IX £p 




allyl acetate 



vinyl acetate 



benzyl acetate 



phenyl acetate 



We shall review all the organic element element symbols you have met at the end of the chapter. 



Names for more complicated but still well known molecules 

Complicated molecules that have been isolated from natural sources are always 
given trivial names, because in these cases, the systematic names really are 
impossible! 

Strychnine is a famous poison featured in many detective stories and a molecule 
with a beautiful structure. All chemists refer to it as strychnine as the systematic name 
is virtually unpronounceable. Two groups of experts at IUPAC and Chemical 
Abstracts also have different ideas on the sys- 
tematic name for strychnine. Others like this are 
penicillin, DNA, and folic acid. 

But the champion is vitamin B 12 , a com- 
plicated cobalt complex with a three-dimen- 
sional structure of great intricacy. No 
chemist would learn this structure but would 
look it up in an advanced textbook of organic 
chemistry. You will find it in such books in 
the index under vitamin Bj2 and not under 
its systematic name. We do not even know 
what its systematic name might be and we 
are not very interested. This is vitamin Bj2- 

Even fairly simple but important mole- 
cules, the amino acids for example, that have 
systematic names that are relatively easy to 
understand are normally referred to by their 



°^/ NH 2 




strychnine, or 
(lR,llR,18S,20S,21S,22S)-12-oxa-8.17- 

diazaheptacycloIlS.S.O 1 ' 8 ^ 2 ' 7 ^ 15 - 20 ] 

tetracosa-2,4,6,14-tetraene-9-one (IUPAC) 

or 

4aR-[4aa,5aa,8aR*,15aa,15ba,15c|3]- 

2,4a,5,5a,7,8,15,15a,15b,15c-decahydro- 

4,6-methano-6H,14H-indolo[3,2,l-/)]oxepino 

[2,3,4-de]pyrrolo[2,3-/i]quinolone 

{Chemical Abstracts) 




vitamin B 12 , or. 



42 



2 ■ Organic structures 



trivial names which are, with a bit of practice, easy to remember and hard to muddle up. They are 
given in full in Chapter 49. 

NH 2 , NH 2 NH 2 



C0 2 H 

alanine, or 
2-aminopropanoic acid 




C0 2 H 



leucine, or 
2-amino-4-methylpentanoic acid 



H 2 N' 



v C0 2 H 



lysine, or 
2,6-diaminohexanoic acid 




toluene 



N0 2 

2,4,6-trinitrotoluene 



A very flexible way of getting new, simple names for compounds can be to combine a bit of sys- 
tematic nomenclature with trivial nomenclature. 
C0 2 H Alanine is a simple amino acid that occurs in proteins. Add a phenyl group and you have pheny- 
lalanine a more complex amino acid also in proteins. 

Toluene, the common name for methylbenzene, can be combined (both chemically and in mak- 
ing names for compounds!) with three nitro groups to give the famous explosive trinitrotoluene or 
TNT. 

Compounds named as acronyms 

Some compounds are referred to by acronyms, shortened versions of either their systematic or their triv- 
ial name. We just saw TNT as an abbreviation for TriNitroToluene but the commoner use for acronyms 
is to define solvents and reagents in use all the time. Later in the book you will meet these solvents. 

Me 



o 



Me' 



Y 



Me' 



N Me 



The names and structures of 
these common solvents need 
learning too. 



Me 



Me 



Me 



Me 



LDA 
Lithium Di-isopropylAmide 



THF 
(TetraHydroFuran) 



DMF 
(DiMethylFormamide) 



DMSO 

(DiMethylSulfOxide) 



The following reagents are usually referred to by acronym and their functions will be introduced 
in other chapters so you do not need to learn them now. You may notice that some acronyms refer to 
trivial and some to systematic names. There is a glossary of acronyms for solvents, reagents, and 
other compounds on p. 000. 



Me 



Me 



Me 



Me 



DIBAL 
Di-lsoButylALuminiumhydride 




V 



CI' 



,0 



PCC 
Pyridinium ChloroChromate 



^N. /C0 2 Et 
Et0 2 (T ^N 



DEAD 
DiEthyl Azo-Dicarboxylate 



Compounds for which chemists use systematic names 

You may be surprised to hear that practising organic chemists use systematic names at all in view of 
what we have just described, but they do! Systematic names really begin with derivatives of pentane 
(C5H12) since the prefix pent- means five, whereas but- does not mean four. Chemists refer to sim- 
ple derivatives of open chain and cyclic compounds with 5 to about 20 carbon atoms by their system- 
atic names, providing that there is no common name in use. Here are some examples. 



r\ 






OH 



cyclopentadiene 



cyclo-octa-l,5-diene 



cyclododeca-l,5,9-triene 



HO 



2,7-dimethyl-3,5-octadiyne-2,7-diol 



How should you name compounds? 43 



„C0 2 H 

/ \^- \/ ^^ -- CH0 

11-bromo-undecanoic acid non-2-enal 



These names contain a syllable that tells you the framework size: penta- for C5, octa- for Cg, nona- 
for C9, undeca- for Cn, and dodeca- for Ci2- These names are easily worked out from the structures 
and, what is more important, you get a clear idea of the structure from the name. One of them might 
make you stop and think a bit (which one?), but the others are clear even when heard without a dia- 
gram to look at. 

Complicated molecules with no trivial names 

When chemists make complex new compounds in the laboratory, they publish them in a chemical 
journal giving their full systematic names in the experimental account, however long and clumsy 
those names may be. But in the text of the paper, and while talking in the lab about the compounds 
they have made, they will just call them 'the amine' or 'the alkene'. Everyone knows which amine or 
alkene is meant because at some point they remember seeing a chemical structure of the compound. 
This is the best strategy for talking about almost any molecule: draw a structure, then give the com- 
pound a 'tag' name like 'the amine' or 'the acid'. In written chemistry it's often easiest to give every 
chemical structure a 'tag' number as well. 

To illustrate what we mean, let's talk about this compound. 

Me Me 
,0^ 




This carboxylic acid was made and used as an intermediate when chemists in California made breve- 
toxin (see p. 000) in 1995. Notice how we can call a complicated molecule 'this acid' — a 'tag' 
name — because you've seen the structure. It also has a tag number (19), so we can also call it 'com- 
pound 19', or 'acid 19', or 'brevetoxin fragment 19'. How much more sensible than trying to work 
out its systematic name. 



How should you name compounds? 



So what should you call a compound? It really depends on circumstances, but you won't go far 
wrong if you follow the example of this book. We shall use the names for compounds that real 



► Our advice on chemical names — six points in order of importance 

• Draw a structure first and worry about the name afterwards 

• Learn the names of the functional groups fester, nitrile, etc.) 

• Learn and use the names of a few simple compounds used by all chemists 

• In speech, refer to compounds as 'that acid' (or whatever) while pointing to a 
diagram 

• Grasp the principles of systematic (IUPAC) nomenclature and use it for com- 
pounds of medium size 

• Keep a notebook to record acronyms, trivial names, structures, etc. that you 
might need later 



44 



2 ■ Organic structures 



chemists use. There's no need to learn all the commonly used names for compounds now, but you 
should log them in your memory as you come across them. Never allow yourself to pass a compound 
name by unless you are sure you know what chemical structure it refers to. You will find many of the 
commonly used names for compounds on the endpapers of this book. Refer to these, or to the short- 
er glossary on p. 000 to refresh your memory should you ever need to. 

We've met a great many molecules in this chapter. Most of them were just there to illustrate 
points so don't learn their structures! Instead, learn to recognize the names of the functional groups 
they contain. However, there were 10 names for simple compounds and three for common solvents 
that we advised you to learn. Cover up the structures on the rest of this page and draw the structures 
for these 13 compounds. 



Important structures to learn 



acetone 



x 

Me^ Me 



toluene 




Me 



ether or 
diethyl ether 



pyridine 








x 






acetaldehyde 




rr 0H 






x 


phenol 


k^ 


formic acid 




Q 


acetic acid 




x 


THF 
(tetrahydrofuran) 


or AcOH 




Me 


benzene 





DMFor 

(dimethylformamide) 

Me 2 NCHO 


1 

N H 
Me^ ^rf 




ethyl acetate 
or EtOAc 




x 


DMSO 
(dimethylsulfoxide) 




II 

Me^ Me 



That's all we'll say on the subject of nomenclature — you'll find that as you practise using these names 
and start hearing other people referring to compounds by name you'll soon pick up the most impor- 
tant ones. But, to reiterate, make sure you never pass a compound name by without being absolutely 
sure what it refers to — draw a structure to check. 



Problems 



45 



• Review box: Table of fragment names 


and organic elements 




R alkyl 




f-Bu 


fert-butyl 


*l< 


Me methyl 


A;h 3 


Ar 


aryl 


any aromatic ring 


Et ethyl 


P^ 


Ph 


phenyl 


„Q 


Pr (or n-Pr) propyl 


^ 


Bn 


benzyl 





Bu (or n-Bu) butyl 


X^-v/- 


Ac 


acetyl 


^ 


APr isopropyl 


-K 




vinyl 


?^ 


ABu isobutyl 


<J\ 




allyl 


^o^- 


s-Bu sec-butyl 


p<Y^\ 


X 


halide 


F, CI, Br, or 1 





Problems 

1. Draw good diagrams of saturated hydrocarbons with seven 
carbon atoms having (a) linear, (b) branched, and (c) cyclic 
frameworks. Draw molecules based on each framework having 
both ketone and carboxylic acid functional groups. 

2. Study the structure of brevetoxin on p. 000. Make a list of the 
different types of functional group (you already know that there 
are many ethers) and of the numbers of rings of different sizes. 
Finally study the carbon framework — is it linear, cyclic, or 
branched? 

3. What is wrong with these structures? Suggest better ways of 
representing these molecules. 

H 
I II 
H— C— C— NH 
I I 

H H— C— H 



Me 



OH 



I 
CH 2 — N— CH 2 



CH 



'/ s 



NH, 



-CH, 



4. Draw structures corresponding to these names. In each case 
suggest alternative names that might convey the structure more 
clearly to someone who is listening to you speak. 

(a) 1 ,4-di- 1 ( 1 -dimethylethyl)benzene 

(b) 2-(prop-2-enyloxy)prop-l-ene 

(c) cyclohexa-l,3,5-triene 



5. Draw one possible structure for each of these molecules, 
selecting any group of your choice for the 'wild card' substituents. 



L 



Ar 



Ar 2 



6. Translate these very poor 'diagrams' of molecules into more 
realistic structures. Try to get the angles about right and, whatev- 
er you do, don't include any square coplanar carbon atoms or 
other bond angles of 90°! 

C 6 H S CH(0H).(CH 2 )4C0C 2 H 5 

0(CH 2 CH 2 ) 2 

(CH 3 0) 2 CHCH=CHCH(OMe) 2 

7. Suggest at least six different structures that would fit the for- 
mula C4H7NO. Make good realistic diagrams of each one and say 
which functional group(s) are present. 

8. Draw and name a structure corresponding to each of these 
descriptions. 

(a) An aromatic compound containing one benzene ring with 
the following substituents: two chlorine atoms having a para 
relationship, a nitro group having an ortho relationship to one of 
the chlorine atoms, and an acetyl group having a meta relation- 
ship to the nitro group. 



46 



2 ■ Organic structures 



(b) An alkyne having a trifluoromethyl substituent at one end 
and a chain of three carbon atoms at the other with a hydroxyl 
group on the first atom, an amino group on the second, and the 
third being a carboxyl group. 

9. Draw full structures for these compounds, displaying the 
hydrocarbon framework clearly and showing all the bonds pre- 
sent in the functional groups. Name the functional groups. 

AcO(CH 2 ) 3 N0 2 

Me0 2 C.CH 2 .0C0Et 

CH 2 =CH.CO.NH(CH 2 ) 2 CN 



10. Identify the oxidation level of each of the carbon atoms in 
these structures with some sort of justification. 



MeN 







CI 



11. If you have not already done so, complete the exercises on pp. 
000 (drawing amino acids) and 000 (giving structures for the 10 
common compounds and three common solvents). 



Determining organic structures 



3 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• What sorts of structure organic 


• Determining structure by X-ray 


• 1 H NMR spectroscopy chll 


molecules have ch2 


crystallography 


• Solving unknown structures 




• Determining structure by mass 


spectroscopically chl5 




spectrometry 






• Determining structure by 13 C NMR 






spectroscopy 






• Determining structure by infrared 






spectroscopy 





Introduction 

Organic structures can be determined accurately and quickly by spectroscopy 

Having urged you, in the last chapter, to draw structures realistically, we now need to answer the 
question: what is realistic? How do we know what structures molecules actually have? Make no mis- 
take about this important point: we really do know what shape molecules have. You wouldn't be far 
wrong if you said that the single most important development in organic chemistry in modern times 
is just this certainty, as well as the speed with which we can be certain. What has caused this revolu- 
tion can be stated in a word — spectroscopy. 



i What is spectroscopy? 
Rays or waves interact with molecules: 

• X-rays are scattered 

• Radio waves make nuclei resonate 

• Infrared waves are absorbed 



Spectroscopy: 

• measures these interactions 

• plots charts of absorption 

• relates interactions with structure 



X-rays give bond lengths and angles. Nuclear magnetic resonance tells us about 
the carbon skeleton of the molecule. Infrared spectroscopy tells us about the types 
of bond in a molecule. 



Structure of the chapter 

We shall first consider structure determination as a whole and then introduce three different 
methods: 

• Mass spectrometry (to determine mass of molecule and atomic composition) 

• Nuclear magnetic resonance (NMR) spectroscopy (to determine carbon skeleton of molecule) 

• Infrared spectroscopy (to determine functional groups in molecule) 

Of these, NMR is more important than all the rest put together and so we shall return to it in Chapter 
11. Then in Chapter 15, after we've discussed a wider range of molecules, there will be a review 
chapter to bring the ideas together and show you how unknown structures are really determined. If 



48 



3 ■ Determining organic structures 



you would like more details of any of the spectroscopic methods we discuss, you should refer to a 
specialized book. 

X-ray is the final appeal 

In Chapter 2 we suggested you draw saturated carbon chains as zig-zags and not in straight lines with 
90° or 180° bond angles. This is because we know they are zig-zags. The X-ray crystal structure of the 
'straight' chain diacid, hexanedioic acid, is shown below. You can clearly see the zig-zag chain the 
planar carboxylic acid groups, and even the hydrogen atoms coming towards you and going away 
from you. It obviously makes sense to draw this molecule realistically as in the second drawing. 



X-ray crystal structures are 
determined by allowing a sample 
of a crystalline compound to 
diffract X-rays. From the resulting 
diffraction pattern, it is possible 
to deduce the precise spatial 
arrangement of the atoms in the 
molecule — except, usually, the 
hydrogen atoms, which are too 
light to diffract the X-rays and 
whose position must be inferred 
from the rest of the structure. 



H0 2 C — (CH 2 )4 — C0 2 H 

hexanedioic acid 











data for structure taken from Cambridge Crystallographic Data Centre 



shape of hexanedioic acid 



Coenzymes are small molecules 
that work hand-in-hand with 
enzymes to catalyse a 
biochemical reaction. 



If you like systematic names, you can 
call methoxatin 4,5-dihydro-4,5-dioxo- 
lH-pyrrolo[2,3-f]quinoline-2,7,9- 
tricarboxylic acid. But you may feel, like 
us, that 'methoxatin' and a diagram or 
the tag name 'the tricarboxylic acid' are 
better. 



This is one question that X-ray answers better than any other method: what shape does a mole- 
cule have? Another important problem it can solve is the structure of a new unknown compound. 
There are bacteria in oil wells, for example, that use methane as an energy source. It is amazing that 
bacteria manage to convert methane into anything useful, and, of course, chemists really wanted to 
know how they did it. Then in 1979 it was found that the bacteria use a coenzyme, given the trivial 
name 'methoxatin', to oxidize methane to methanol. Methoxatin was a new compound with an 
unknown structure and could be obtained in only very small amounts. It proved exceptionally diffi- 
cult to solve the structure by NMR but eventually methoxatin was found by X-ray crystallography to 
be a polycyclic tricarboxylic acid. This is a more complex molecule than hexanedioic acid but X-ray 
crystallographers routinely solve much more complex structures than this. 




0(4) 



\ 



#0(1) 



CI2) C^) 

\. C(l)/ , C(9) 



C(4) 



yc (12) \-«a, 

0(3) / 



C(7) 



C(ll) 



1 



N(2) 



0(6) 



c 

0(8) (J 



•° 



C(14) 



C(5) 
C(6) . I 0(13) 



6 0(5) 



data for the X-ray structure taken from the Cambridge Crystallographic 
Data Centre 

X-ray crystallography has its limitations 

If X-ray crystallography is so powerful, why do we bother with other methods? There are two reasons. 

• X-ray crystallography works by the scattering of X-rays from electrons and requires crystalline 
solids. If an organic compound is a liquid or is a solid but does not form good crystals, its struc- 
ture cannot be determined in this way. 



Introduction 



49 



• X-ray crystallography is a science in its own right, a separate discipline from chemistry because it 
requires specific skills, and a structure determination can take a long time. Modern methods have 
reduced this time to a matter of hours or less, but nonetheless by contrast a modern NMR 
machine with a robot attachment can run more than 100 spectra in an overnight run. So we nor- 
mally use NMR routinely and reserve X-rays for difficult unknown structures and for determin- 
ing the detailed shape of important molecules. 

Outline of structure determination by spectroscopy 

Put yourself in these situations. 

• Finding an unknown product from a chemical reaction 

• Discovering an unknown compound from Nature 

• Detecting a suspected food contaminant 

• Routinely checking purity during the manufacture of a drug 

In all cases except perhaps the second you need a quick and reliable answer. Suppose you are trying 
to identify the heart drug propranolol, one of the famous 'beta blockers' used to reduce high blood 
pressure and prevent heart attacks. You would first want to know the molecular weight and atomic 
composition and this would come from a mass spectrum: propranolol has a molecular weight (rela- 
tive molecular mass) of 259 and the composition C16H21NO2. Next you would need the carbon 
skeleton — this would come from NMR, which would reveal the three fragments shown. 

fragments of propranolol 
from the NMR spectrum 




propranolol C16H21NO2 




There are many ways in which these fragments could be joined together and at this stage you 
would have no idea whether the oxygen atoms were present as OH groups or as ethers, whether the 
nitrogen would be an amine or not, and whether Y and Z might or might not be the same atom, say 
N. More information comes from the infrared spectrum, which highlights the functional groups, 
and which would show that there is an OH and an NH in the molecule but not functional groups 
such as CN or NO2. This still leaves a variety of possible structures, and these could finally be distin- 
guished by another technique, H NMR. We are in fact going to avoid using H NMR in this chapter, 
because it is more difficult, but you will learn just how much information can be gained from mass 
spectra, IR spectra, and C NMR spectra. 

Now we must go through each of these methods and see how they give the information they do. 
For this exercise, we will use some compounds you encounter in everyday life, perhaps without real- 
izing it. 



H NMR makes an entrance in 
Chapter 11. 



50 



3 ■ Determining organic structures 



Mass spectrometry uses a 
different principle from the other 
forms of spectroscopy we 
discuss: what is measured is not 
absorption of energy but the mass 
of the molecule or fragments of 
molecule. 



• What each spectroscopic method tells us 


Method and what it does 


What it tells us 


Type of data provided 


Mass spectrum weighs the molecule 


molecular weight (relative 
molecular mass) and composition 


259;C 16 H 2 iN0 2 


13 C NMR reveals all different carbon 


carbon skeleton 


no C=0 group; ten carbons 


nuclei 




in aromatic rings; two 
carbons next to 0; three 
other saturated C atoms 


Infrared reveals chemical bonds 


functional groups 


no C=0 group; one OH; 
one NH 





Mass spectrometry 

Mass spectrometry weighs the molecule 

A mass spectrometer has three basic components: something to volatilize and ionize the molecule 
into a beam of charged particles; something to focus the beam so that particles of the same 
massxharge ratio are separated from all others; and something to detect the particles. All spectrome- 
ters in common use operate in a high vacuum and usually use positive ions. Two methods are used to 
convert neutral molecules into cations: electron impact and chemical ionization. 




'0 © 



loss of one electron 
leaves a radical cation 



Mass spectrometry by electron impact 

In electron impact (E.I.) mass spectrometry the molecule is bombarded with highly energetic elec- 
trons that knock a weakly bound electron out of the molecule. If you think this is strange, think of 
throwing bricks at a brick wall: the bricks do not stick to the wall but knock loose bricks off the top of 
the wall. Losing a single electron leaves behind a radical cation: an unpaired electron and a positive 
charge. The electron that is lost will be one of relatively high energy (the bricks come from the top of 
the wall), and this will typically be one not involved in bonding, for example, an electron from a lone 
pair. Thus ammonia gives NH3" and a ketone gives R2C=0 + '. If the electron beam is not too high in 
energy, some of these rather unstable radical cations will survive the focusing operation and get to the 
detector. Normally two focusing operations are used: the beam is bent magnetically and electrostati- 
cally to accelerate the cations on their way to the detector and it takes about 20 (is for the cations to get 
there. But if, as is often the case, the electron beam supplies more than exactly the right amount of 
energy to knock out the electron, the excess energy is dissipated by fragmentation of the radical cation. 
Schematically, an unknown molecule first forms the radical cation M + " which then breaks up (frag- 
ments) to give a radical X' and a cation Y + . Only charged particles (cations in most machines) can be 
accelerated and focused by the magnetic and electrostatic fields and so the detector records only the 
molecular ion M + ' and positively charged fragments Y + . Uncharged radicals X' are not recorded. 



Mass spectrometry 



51 



®o 



electron 
bombardment 



charged ,-. detectable 
+ 



charged .-. detectable 



fragmentation 



unknown molecule with 
a lone pair of electrons 



molecule has lost 
one electron and is 
now a radical cation 







uncharged .-. not detectable 



A typical result is the E.I. mass spectrum for the alarm pheromone of the honey bee. The bees 
check every insect coming into the hive for strangers. If a strange insect (even a bee from another 
hive) is detected, an alarm pheromone is released and the intruder is attacked. The pheromone is a 
simple volatile organic molecule having this mass spectrum. 

mass spectrum of honey bee alarm pheromone 



100- 



80- 



8 60- 



40- 



20- 



43 



55 



58 



59 



71 



99 



114. 



40 



80 



120 



m/z 



Insects communicate by releasing 
compounds with strong smells (to the 
insect!). These have to be small 
volatile molecules, and those used to 
communicate between members of the 
same species are called pheromones. 



The strongest peak, at 43 mass units in this case, is assigned an 'abundance' of 100% and called the 
base peak. The abundance of the other peaks is shown relative to the base peak. In this spectrum, 
there is only one other strong peak (58 at 50%) and the peak of highest mass at 114 (at 5%) is the 
molecular ion corresponding to a structure CjH^O. The main fragmentation is to a CsH^ radical 
(not observed as it isn't charged) and a cation C2H 3 + , which forms the base peak. The pheromone 
is the simple ketone heptan-2-one. 




electron 
bombardment 

»- 



O 




fragmentation 



heptan-2-one 
(only one lone pair shown) 



M + * = 114 = C 7 H 14 



: CrHi 



71 



Mass spectroscopy requires 
minute quantities of sample — 
much less than the amounts 
needed for the other techniques 
we will cover. Pheromones are 
obtainable from insects only on a 
microgram scale or less. 



u 

[I 



Y® =C 2 H 5 0=43 



The problem with E.I. is that for many radical cations even 20 (is is too long, and all the molecular 
ions have decomposed by the time they reach the detector. The fragments produced may be useful in 
identifying the molecule, but even in the case of the bee alarm pheromone it would obviously be bet- 
ter to get a stronger and more convincing molecular ion as the weak (5%) peak at 1 14 might also be a 
fragment or even an impurity. 



52 3 ■ Determining organic structures 



Mass spectrometry by chemical ionization 

In chemical ionization (C.I.) mass spectrometry the electron beam is used to ionize a simple mole- 
cule such as methane which in turn ionizes our molecule by collision and transfer of a proton. Under 
electron bombardment, methane loses a bonding electron (it doesn't have any other kind) to give 
CH4 ' which reacts with an unionized methane molecule to give CH3 and CH5 . Before you write in 
complaining about a mistake, just consider that last structure in a bit more detail. Yes, CH5 does 
have a carbon atom with five bonds. But it has only eight electrons! These are distributed between 
five bonds (hence the + charge) and the structure is thought to be trigonal bipyramidal. This struc- 
ture has not been determined as it is too unstable. It is merely proposed from theoretical calculations. 



electron lost under ', 

electron bombardment .a* ^"4 . ©; 

CH 4 »- CH 4 »- CH 3 + H — C! 



H 



© 
proposed structure of CH 5 

the two black bonds share two electrons 

This unstable compound is a powerful acid, and can protonate just about any other molecule. 
When it protonates our sample, a proton has been added rather than an electron removed, so the 
resulting particles are simple cations, not radical cations, and are generally more stable than the rad- 
ical cations produced by direct electron impact. So the molecular ion has a better chance of lasting 
the necessary 20 (is to reach the detector. Note that we now observe [M + H] + (i.e. one more than the 
molecular mass) rather than M + by this method. 

Having more functional groups helps molecular ions to decompose. The aromatic amine 
2-phenylethylamine is a brain active amine found in some foods such as chocolate, red wine, and 
cheese and possibly implicated in migraine. It gives a poor molecular ion by E.I., a base peak with a 
mass as low as 30 and the only peak at higher mass is a 15% peak at 91. The C.I. mass spectrum on the 
other hand has a good molecular ion: it is [M + H] + of course. Normally a fragmentation gives one 
cation and another radical, only the cation being detected. It is relatively unusual for one bond to be 
able to fragment in either direction, but here it does, which means that both fragments are seen in the 
spectrum. 

the radical cation can fragment in two ways 

30 M + * very small 91 

uncharged radical - _ ^21 = C H N uncharged radical - 

not seen not seen 

Mass spectrometry separates isotopes 

You will know in theory that most elements naturally exist as mixtures of isotopes. If you didn't 
believe it, now you will. Chlorine is normally a 3:1 mixture of CI and CI (hence the obviously 
false relative atomic mass of '35.5' for chlorine) while bromine is an almost 1:1 mixture of Br and 
Br (hence the 'average' mass of 80 for bromine!). Mass spectrometry separates these isotopes so 
that you get true not average molecular weights. The molecular ion in the E.I. mass spectrum of the 
bromo-amide below has two peaks at 213 and 215 of roughly equal intensity. This might just repre- 
sent the loss of molecular hydrogen from a molecular ion 215, but, when we notice that the first frag- 
ment (and base peak) has the same pattern at 171/173, the presence of bromine is a more likely 
explanation. All the smaller fragments at 155, 92, etc. lack the 1:1 pattern and also therefore lack 
bromine. 





Mass spectrometry 



53 



100- 

80- 
60- 
40- 
20- 

0. 
100- 
80- 
60- 
40- 
20- 

0- 



74.0 81.0 



92.1 

98.0. 105.5 




117.0 12 f° 132.0 

■■i i ■■ i " i I !'i i i- I t v 



143.0 
,148.0 



120 



140 



213.0 



198.9 

' i-i'l r ' 
200 



224.0 



220 



-+- 



236.0 

i I I T 



180 



240 




fragmentation 





M + 213/215 



171/173 



92 



Table 3.1 Summary table of main isotopes for mass spectra 
Element Carbon Chlorine Bromine 



isotopes 
rough ratio 



12 C, 13 C 



35q| 37 



CI 



Br 



The mass spectrum of chlorobenzene (PhCl, C6H5CI) is very simple. There are two peaks at 112 
(100%) and 114 (33%), a peak at 77 (40%), and very little else. The peaks at 112/114 with their 3:1 
ratio are the molecular ions, while the fragment at 77 is the phenyl cation (Ph + or C6H5 ). 

The mass spectrum of DDT 

is very revealing. This very 

effective insecticide became 

notorious as it accumulated in 

the fat of birds of prey (and 

humans) and was phased out 

of use. It can be detected easily 

by mass spectrometry because the five chlorine atoms produce a complex molecular ion at 

252/254/256/258/260 with ratios of 243:405:270:90:15:1 (the last is too small to see). The peak at 252 

contains nothing but CI, the peak at 254 has four atoms of CI and one atom of CI, while the 

invisible peak at 260 has five CI atoms. The ratios need some working out, but the first fragment at 

235/237/239 in a ratio 9:6:1 is easier. It shows just two chlorine atoms as the CCI3 group has been lost 

as a radical. 



It's worth remembering that the 
Ph + weighs 77: you'll see this 
mass frequently. 



1.1% 13 C (90:1) 3:1 



79 Br 81 



1:1 



Remember: mass spectroscopy is 
very good at detecting minute 
quantities. 



54 



3 ■ Determining organic structures 



Isotopes in DDT 

The ratio comes from the 3:1 isotopic ratio like this: 

• chance of one 35 CI inthe molecule: — 

4 

•chanceofone CI inthe molecule: — 

4 

If the molecule or fragment contains two chlorine atoms, as does our Ci3HgCl2, then 

• chance of two 35 Cls inthe molecule: — x — = — 



|_4 X 4_| + |_4 X 4_T 



_6_ 
16 



• chance of one 35 CI and one 37 CI in the molecule: 

• chance of two 37 Cls inthe molecule: —x — = — 

4 4 16 

The ratio of these three fractions is 9:6:1, the ratio of the peaks inthe mass spectrum. 

mass spectrum of DDT 
100- 



80- 



60- 



40- 



20- 



ecu 




,.i ..I J .l. | l,il.,,l|lll.iLii | i 1. ..il. r l Jl... .ILllll lit (|L |||| 



100 



200 

m/z 



300 



400 




fragmentation 



CCI 3 + 

stable CI 

radical 




CI 



stable cation 



"~ 1" denotes the cation radical produced 
by E.I. 



■I -J 

Carbon has a minor but important isotope C 

Many elements have minor isotopes at below the 1% level and we can ignore these. One important 
one we cannot ignore is the 1 . 1 % of C present in ordinary carbon. The main isotope is C and you 
may recall that C is radioactive and used in carbon dating, but its natural abundance is minute. 
The stable isotope C is not radioactive, but it is NMR active as we shall soon see. If you look back at 
the mass spectra illustrated so far in this chapter, you will see a small peak one mass unit higher than 
each peak in most of the spectra. This is no instrumental aberration: these are genuine peaks con- 
taining C instead of C. The exact height of these peaks is useful as an indication of the number of 
carbon atoms in the molecule. If there are n carbon atoms in a molecular ion, then the ratio of M + to 
[M+ l] + isl00:(l.l)x n. 



Mass spectrometry 



55 



The electron impact mass spectrum of BHT gives a good example. The molecular ion at 220 has 
an abundance of 34% and [M + 1] + at 221 has 5-6% abundance but is difficult to measure as it is so 
weak. BHT is Ci 5 H 2 60 so this should give an [M + 1] + peak due to 13 C of 15 x 1.1% of M + , that is, 
16.5% of M + or 34 x 16.5 = 5.6% actual abundance. An easier peak to interpret is the base peak at 
205 formed by the loss of one of the six identical methyl groups from the f-butyl side chains (don't 
forget what we told you in Chapter 2 — all the 'sticks' in these structures are methyl groups and not 
hydrogen atoms). The base peak (100%) 205 is [M — Me] + and the 13 C peak 206 is 15%, which fits 
well with 14 x 1.1% = 15.4%. 



mass spectrum of BHT 



100 



80- 



60 



ra 40- 

"5 



20 




Id, J|ii Jiilliyiiiili, Lip Jul, .iiilii,,, IiiI|L illililii.^iiiliiii, \y„ r ,ili Jj 



UL ^ 



50 



100 



150 



200 



250 



BHT 

BHT is used to prevent the 
oxidation of vitamins A and E in 
foods. It carries the E-number 
E321. There has been some 
controversy over its use because it 
is a cancer suspect agent, but it is 
used in some 'foods' like chewing 
gum. BHT stands for 'Butylated 
HydroxyToluene', but you can call it 
2,6-di-f-butyl-4-methylphenol if you 
want to, but you maypreferto look 
at the structure and just call it BHT. 
You met BHT briefly in Chapter 2 
when you were introduced to the 
tertiary butyl group. 



m/z 




'BHT' 
C15H26O 




C-14H03O 



Other examples you have seen include the DDT spectrum, where the peaks between the main 
peaks are C peaks: thus 236, 238, and 240 are each 14% of the peak one mass unit less, as this frag- 
ment has 13 carbon atoms. If the number of carbons gets very large, so does the C peak; eventually 

it is more likely that the molecule contains one C than that it 

doesn't. We can ignore the possibility of two C atoms as 1.1% of 
1.1% is very small (probability of 1.32 x 10~ 5 ). 

Table 3.2 summarizes the abundance of the isotopes in these 
three elements. Notice that the ratio for chlorine is not exactly 3:1 
nor that for bromine exactly 1:1; nevertheless you should use the 
simpler ratios when examining a mass spectrum. Always look at the 
heaviest peak first: see whether there is chlorine or bromine in it, 



Table 3. 2 Abundance of isotopes for carbon, chlorine, and bromine 



Element 

carbon 

chlorine 

bromine 



Major isotope: abundance 

12 C: 98.9% 
35 CI: 75.8% 
79 Br: 50.5% 



Minor isotope: abundance 



J C: 1.1% 



37 CI: 24.2% 
81 Br:49.5% 



56 



3 ■ Determining organic structures 



and whether the ratio of M + to [M + 1] + is about right. If, for example, you have what seems to be M + 
at 120 and the peak at 121 is 20% of the supposed M + at 120, then this cannot be a C peak as it would 
mean that the molecule would have to contain 18 carbon atoms and you cannot fit 18 carbon atoms 
into a molecular ion of 120. Maybe 121 is the molecular ion. 

Atomic composition can be determined by high resolution mass spectrometry 

Ordinary mass spectra tell us the molecular weight (MW) of the molecule: we could say that the bee 
alarm pheromone was MW 114. When we said it was C7H44O we could not really speak with 
confidence because 114 could also be many other things such as CgHjg or C6H10O2 or C6H14N2. 
These different atomic compositions for the same molecular weight can nonetheless be distinguished 
if we know the exact molecular weight, since individual isotopes have non-integral masses (except C 
by definition). Table 3.3 gives these to five decimal places, which is the sort of accuracy you need for 
meaningful results. Such accurate mass measurements are called high resolution mass spectrometry. 
For the bee alarm pheromone, the accurate mass turns out to be 114.1039. Table 
3.4 compares possible atomic compositions, and the result is conclusive. The exact 
masses to three places of decimals fit the observed exact mass only for the composition 
CjH^O. You may not think the fit is very good when you look at the two numbers, 
but notice the difference in the error expressed as parts per million. One answer stands 
out from the rest. Note that even two places of decimals would be enough to distin- 
guish these four compositions. 

A more important case is that of the three ions at 28: nitrogen, carbon monoxide, 
and ethylene (ethene, CH2=CH2). Actually mass spectra rarely go down to this low 
value because some nitrogen is usually injected along with the sample, but the three 
ions are all significant and it is helpful to see how different they are. Carbon monoxide 
CO is 27.9949, nitrogen N 2 is 28.0061, and ethylene 28.0313. 



The reason that exact masses are 
not integers lies in the slight 
mass difference between a 
proton (1.67262 x 10" 27 kg) and 
a neutron (1.67493 x 10" 27 kg) 
and in the fact that electrons have 
mass (9.10956 x 10" 31 kg). 



Table 3.3 Exact masses 


of common elements 


Element 

hydrogen 


Isotope 


Atomic 
weight 

1 


Exact mass 

1.00783 


carbon 


12 C 


12 


12.00000 


carbon 


13 C 


13 


13.00335 


nitrogen 


14 N 


14 


14.00307 


oxygen 


16 


16 


15.99492 


fluorine 


19 F 


19 


18.99840 


phosphorus 


31p 


31 


30.97376 


sulfur 


32 s 


32 


31.97207 


chlorine 


35 C | 


35 


34.96886 


chlorine 


37 CI 


37 


36.96590 


bromine 


79 Br 


79 


78.91835 


bromine 


81g r 


81 


80.91635 



In the rest of the book, whenever we state that a molecule has a 
certain atomic composition, you can assume that it has been 
determined by high resolution mass spectrometry on the molecular 



ion. 



Table 3.4 Exact mass determination for the bee alarm 
pheromone 



Composition 


Calculated 


Observed 


Error in 




NT 


NT 


p.p.m. 


C6H10O2 


114.068075 


114.1039 


358 


C5H14N2 


114.115693 


114.1039 


118 


C7H3.4O 


114.104457 


114.1039 


5 


CsHi8 


114.140844 


114.1039 


369 



This rule holds as long as there 
are only C, H, N, 0, S atoms in the 
molecule. It doesn't work for 
molecules with CI or P atoms for 
example. 



One thing you may have noticed in Table 3.4 is that there are no entries 
with just one nitrogen atom. Two nitrogen atoms, yes; one nitrogen no! This 
is because any complete molecule with one nitrogen in it has an odd molecular 
weight. Look back at the mass spectrum of the compounds giving good mole- 
cular ions by C.I. for an example. The nitro compound had M = 127 and the 
amine M = 121. This is because C, O, and N all have even atomic weights — 
only H has an odd atomic weight. Nitrogen is the only element from C, O, 
and N that can form an odd number of bonds (3). Molecules with one nitro- 
gen atom must have an odd number of hydrogen atoms and hence an odd 
molecular weight. Molecules with only C, H, and O or with even numbers of 
nitrogen atoms have even molecular weights. 
If we are talking about fragments, that is, cations or radicals, the opposite applies. A fragment has, 
by definition, an unused valency. Look back at the fragments in this section and you will see that this 
is so. Fragments with C, H, O alone have odd molecular weights, while fragments with one nitrogen 
atom have even molecular weights. 

Nuclear magnetic resonance 

What does it do? 

Nuclear magnetic resonance (NMR) allows us to detect atomic nuclei and say what sort of environ- 
ment they are in, within their molecule. Clearly, the hydrogen of, say, propanol's hydroxyl group is 



Nuclear magnetic resonance 



57 



H 



different from the hydrogens of its carbon skeleton — it can be displaced by sodium metal, for exam- 
ple. NMR (actually H, or proton, NMR) can easily distinguish between these two sorts of hydro- 
gens. Moreover, it can also distinguish between all the other different sorts of hydrogen atoms 
present. Likewise, carbon (or rather C) NMR can easily distinguish between the three different car- 
bon atoms. In this chapter we shall look at C NMR spectra and then in Chapter 1 1 we shall look at 
proton ( 1 H) NMR spectra in detail. 

NMR is incredibly versatile: it can even scan living human brains (see picture) but the principle is 
still the same: being able to detect nuclei (and hence atoms) in different environments. We need first 
to spend some time explaining the principles of NMR. 





H I A 



Proton NMR can distinguish 
between the different coloured 
hydrogens. Carbon NMR can 
distinguish between all the 
carbons. 



When NMR is used medically it is 
usually called Magnetic Resonance 
Imaging (MRI) for fear of frightening 
patients wary of all things nuclear. 



NMR uses a strong magnetic field 

Imagine for a moment that we were able to 'switch off the earth's magnetic field. One effect would be 
to make navigation much harder since all compasses would be useless. They would be free to point in 
whatever direction they wanted to and, if we turned the needle round, it would simply stay where we 
left it. However, as soon as we switched the magnetic field back on, they would all point north — their 
lowest energy state. Now if we wanted to force a needle to point south we would have to use up energy 
and, of course, as soon as we let go, the needle would return to its lowest energy state, pointing north. 

In a similar way, some atomic nuclei act like tiny compass needles and have different energy levels 
when placed in a magnetic field. The compass needle can rotate through 360° and have an essentially 
infinite number of different energy levels, all higher in energy than the 'ground state' (pointing 
north). Fortunately, our atomic nucleus is more restricted — its energy levels are quantized, just like 
the energy levels of an electron, which you will meet in the next chapter, and there are only certain 
specific energy levels it can adopt. This is like allowing our compass needle to point, say, only north 
or south. Some nuclei (including 'normal' carbon- 12) do not interact with a magnetic field at all and 
cannot be observed in an NMR machine. The nuclei we shall be looking at, H and C, do interact 
and have just two different energy levels. When we apply a magnetic field to these nuclei, they can 
either align themselves with it, which would be the lowest energy state, or they can align themselves 
against the field, which is higher in energy. 

Let us return to the compass for a moment. We have already seen that if we could switch off the 
earth's magnetic field it would be easy to turn the compass needle round. When it is back on we need 
to push the needle (do work) to displace it from north. If we turned up the earth's magnetic field still 
more, it would be even harder to displace the compass needle. Exactly how hard it is to turn the com- 
pass needle depends on how strong the earth's magnetic field is and also on how well our needle is 
magnetized — if it is only weakly magnetized, it is much easier to turn it round and, if it isn't magne- 
tized at all, it is free to rotate. 

Likewise, with our nucleus in a magnetic field, the difference in energy between the nuclear spin 
aligned with and against the applied field depends on how strong the magnetic field is, and also on 
the properties of the nucleus itself. The stronger the magnetic field we put our nucleus in, the greater 
the energy difference between the two alignments. Now here is an unfortunate thing about NMR: 
the energy difference between the nuclear spin being aligned with the magnetic field and against it is 
really very small — so small that we need a very, very strong magnetic field to see any difference at all. 



Nuclei that interact with magnetic 
fields are said to possess nuclear 
spin. The exact number of 
different energy levels a nucleus 
can adopt is determined by this 
nuclear spin, /, of the particular 
isotope. The nuclearspin /can 
have various values such as 0, |, 
1, | and the number of energy 
levels is given by 2/+ 1. Some 
examples are: 1 H, / = |; 2 H (= D), 

;=o. 



/=1; J - L B,/ 



. 5. 12 r 
2' 



58 3 ■ Determining organic structures 



NMR machines contain very strong electromagnets 

The earth's magnetic field has a field strength of 2 x 1CT 5 have become firmly attached to NMR magnets. Even with the 

tesla. A typical magnet used in an NMR machine has a field immensely powerful magnets used the energy difference is 

strength of between 2 and 10 tesla, some 10 5 times stronger stil1 s0 smal1 that the nuclei only have a very small preference 

than the earth's field. These magnets are dangerous and no for the lower energy state. Fortunately, we can just detect this 

metal objects must be taken into the rooms where they are: small preference, 
stories abound of unwitting workmen whose metal toolboxes 



NMR also uses radio waves 

A H or C nucleus in a strong magnetic field can have two energy levels. We could do work to make 
our nucleus align against the field rather than with it (just like turning the compass needle round). 
But since the energy difference between the two states is so small, we don't need to do much work. In 
fact, the amount of energy needed to flip the nucleus can be provided by electromagnetic radiation of 
radio-wave frequency. Radio waves flip the nucleus from the lower energy state to the higher state. 
The nucleus now wants to return to the lower energy state and, when it does so, the energy comes out 
again and this (a tiny pulse of radio frequency electromagnetic radiation) is what we detect. 
We can now sum up how an NMR machine works. 

1 The sample of the unknown compound is dissolved in a suitable solvent and put in a very strong 
magnetic field. Any atomic nuclei with a nuclear spin now have different energy levels, the exact 
number of different energy levels depending on the value of the nuclear spin. For H and C 
NMR there are two energy levels 

2 The sample is irradiated with a short pulse of radiofrequency energy. This disturbs the equilibri- 
um balance between the two energy levels: some nuclei absorb the energy and are promoted to a 
higher energy level 

3 We then detect the energy given out when the nuclei fall back down to the lower energy level 
using what is basically a sophisticated radio receiver 



< Insert Graphic 03.022 (picture of NMR and diagram of components; not provided) > 



4 After lots of computation, the results are displayed in the form of intensity (i.e. number of 
absorptions) against frequency. Here is an example, which we shall return to in more detail later 



i 



200 150 100 50 

Why do chemically distinct nuclei absorb energy at different frequencies? 

In the spectrum you see above, each line represents a different kind of carbon atom: each one absorbs 
energy (or resonates — hence the term nuclear magnetic resonance) at a different frequency. But why 
should carbon atoms be 'different'? We have told you two factors that affect the energy difference 
(and therefore the frequency) — the magnetic field strength and what sort of nucleus is being studied. 
So you might expect all carbon- 13 nuclei to resonate at one particular frequency and all protons 
( H) to resonate at one (different) frequency. But they don't. 



Nuclear magnetic resonance 



59 



The variation in frequency for different carbon atoms must mean that the energy jump from 
nucleus-aligned-with to nucleus-aligned-against the applied magnetic field must be different for 
each type of carbon atom. The reason there are different types of carbon atom is that their nuclei 
experience a magnetic field that is not quite the same as the magnetic field that we apply. Each nucle- 
us is surrounded by electrons, and in a magnetic field these will set up a tiny electric current. This 
current will set up its own magnetic field (rather like the magnetic field set up by the electrons of an 
electric current moving through a coil of wire or solenoid), which will oppose the magnetic field that 
we apply. The electrons are said to shield the nucleus from the external magnetic field. If the electron 
distribution varies from C atom to C atom, so does the local magnetic field, and so does the res- 
onating frequency of the C nuclei. Now, you will see shortly (in Chapter 5) that a change in elec- 
tron density at a carbon atom also alters the chemistry of that carbon atom. NMR tells us about the 
chemistry of a molecule as well as about its structure. 



shielding of nuclei from an applied magnetic field by electrons: 



applied I 

magnetic I 

field | 




small induced magnetic field 
shielding the nucleus 



i Changes in the distribution of electrons around a nucleus affect: 

• the local magnetic field that the nucleus experiences 

• the frequency at which the nucleus resonates 

• the chemistry of the molecule at that atom 

This variation in frequency is known as the chemical shift. Its symbol is 5. 



As an example, consider ethanol, CH3CH2OH. The carbon attached to the OH group will have 
relatively fewer electrons around it compared to the other carbon since the oxygen atom is more 
electronegative and draws electrons towards it, away from the carbon atom. 

The magnetic field that this (red) carbon nucleus feels will therefore be slightly greater than that 
felt by the (green) carbon with more electrons since the red carbon is less shielded from the applied 
external magnetic field — in other words it is deshielded. Since the carbon attached to the oxygen 
feels a stronger magnetic field, there will be a greater energy difference between the two alignments of 
its nucleus. The greater the energy difference, the higher the resonant frequency. So for ethanol we 
would expect the red carbon with the OH group attached to resonate at a higher frequency than the 
green carbon, and indeed this is exactly what the C NMR spectrum shows. 



H H 

A 

H H 

ethanol 
We have shown all the Cs and Hs here 
because we want to talk about them. 



200 



150 



100 



50 



The peaks at 77 p. p.m., coloured 
brown, are those of the usual 
solvent (CDCI3) and can be 
ignored for the moment. We shall 
explain them in Chapter 15. 



The chemical shift scale 

When you look at an NMR spectrum you will see that the scale does not appear to be in magnetic 
field units, nor in frequency units, but in 'parts per million' (p. p.m.). There is an excellent reason for 



60 



3 ■ Determining organic structures 



this and we need to explain it. The exact frequency at which the nucleus resonates depends on the 
external applied magnetic field. This means that, if the sample is run on a machine with a different 
magnetic field, it will resonate at a different frequency. It would make life very difficult if we couldn't 
say exactly where our signal was, so we say how far it is from some reference sample, as a fraction of 
the operating frequency of the machine. We know that all protons resonate at approximately the 
same frequency in a given magnetic field and that the exact frequency depends on what sort of chem- 
ical environment it is in, which in turn depends on its electrons. This approximate frequency is the 
operating frequency of the machine and simply depends on the strength of the magnet — the stronger 
the magnet, the larger the operating frequency. The precise value of the operating frequency is sim- 
ply the frequency at which a standard reference sample resonates. In everyday use, rather than actu- 
ally referring to the strength of the magnet in tesla, chemists usually just refer to its operating 
frequency. A 9.4 T NMR machine is referred to as a 400 MHz spectrometer since that is the fre- 
quency in this strength field at which the protons in the reference sample resonate; other nuclei, for 
example C, would resonate at a different frequency, but the strength is arbitrarily quoted in terms 
of the proton operating frequency. 

The reference sample — tetramethylsilane, TMS 

The compound we use as a reference sample is usually tetramethylsilane, TMS. This is silane (S1H4) 
with each of the hydrogen atoms replaced by methyl groups to give Si(CH3)4. The four carbon atoms 
attached to silicon are all equivalent and, because silicon is more electropositive than carbon, are 
fairly electron-rich (or shielded), which means they resonate at a frequency a little less than that of 
most organic compounds. This is useful because it means our reference sample is not bang in the 
middle of our spectrum! 

The chemical shift, 5, in parts per million (p. p.m.) of a given nucleus in our sample is defined in 
terms of the resonance frequency as: 



H 3 C^ .^CH 3 

H3C CH3 

tetramethylsilane, TMS 



Silicon and oxygen have opposite 
effects on an adjacent carbon 
atom: silicon shields; oxygen 
deshields. 

Electronegativities: Si: 1.8; C: 
2.5; 0:3.5. 



c frequency (Hz)- frequency TMS (Hz) 

o = 

frequency TMS (MHz) 

No matter what the operating frequency (i.e. strength of the magnet) of the NMR machine, the 
signals in a given sample (e.g. ethanol) will always occur at the same chemical shifts. In ethanol the 
(red) carbon attached to the OH resonates at 57.8 p. p.m. whilst the (green) carbon of the methyl 
group resonates at 18.2 p. p.m. Notice that by definition TMS itself resonates at p. p.m. The carbon 
nuclei in most organic compounds resonate at greater chemical shifts, normally between and 
200 p. p.m. 

Now, let's return to the sample spectrum you saw on p. 000 and which is reproduced below, and 
you can see the features we have discussed. This is a 100 MHz spectrum; the horizontal axis is actual- 
ly frequency but is usually quoted in p. p.m. of the field of the magnet, so each unit is one p. p.m. of 
100 MHz, that is, 100 Hz. We can tell immediately from the three peaks at 176.8, 66.0, and 19.9 
p. p.m. that there are three different types of carbon atom in the molecule. 



Again, ignore the brown solvent 
peaks — they are of no interest to us at 
the moment. You also need not worry 
about the fact that the signals have 
different intensities. This is a 
consequence of the way the spectrum 
was recorded. 



200 



150 



100 



50 



But we can do better than this: we can also work out what sort of chemical environment the car- 
bon atoms are in. All C spectra can be divided into four major regions: saturated carbon atoms 
(0-50 p.p.m.), saturated carbon atoms next to oxygen (50-100 p. p.m.), unsaturated carbon atoms 
(100-150 p.p.m.), and unsaturated carbon atoms next to oxygen, i.e. C=0 groups (150-200 p.p.m.). 



Nuclear magnetic resonance 



61 



Regions of the 13 C NMR spectrum (scale in p. p.m.) 



Unsaturated carbon 
atoms next to 
oxygen (C=0) 

5=200-150 



Unsaturated carbon 
atoms (C=C and 
aromatic carbons) 

5=150-100 



Saturated carbon 
atoms next to oxygen 
(CH 3 0, CH 2 0, etc.) 

5=100-50 



Saturated carbon 
atoms (CH 3 , CH 2 , 
CH) 

5 = 50-0 



The spectrum you just saw is in fact of lactic acid (2-hydroxypropanoic acid). When you turned 
the last page, you made some lactic acid from glucose in the muscles of your arm — it is the break- 
down product from glucose when you do anaerobic exercise. Each of lactic acid's carbon atoms gives 
a peak in a different region of the spectrum. 

Different ways of describing chemical shift 

The chemical shift scale runs to the left from zero (where TMS resonates) — i.e. backwards from the 
usual style. Chemical shift values around zero are obviously small but are confusingly called 'high 
field' because this is the high magnetic field end of the scale. We suggest you say 'large' or 'small' 
chemical shift and 'large' or 'small' 5, but 'high' or 'low' field to avoid confusion. Alternatively, use 
'upfield' for high field (small 5)and 'downfield' for low field (large 5). 

One helpful description we have already used is shielding. Each carbon nucleus is surrounded by 
electrons that shield the nucleus from the applied field. Simple saturated carbon nuclei are the most 
shielded: they have small chemical shifts (0-50 p. p.m.) and resonate at high field. One electro- 
negative oxygen atom moves the chemical shift downfield into the 50-100 p.p.m. region. The 
nucleus has become deshielded. Unsaturated carbon atoms experience even less shielding (100-150 
p.p.m.) because of the way in which electrons are distributed around the nucleus. If the % bond is to 
oxygen, then the nucleus is even more deshielded and moves to the largest chemical shifts around 
200 p.p.m. The next diagram summarizes these different ways of talking about NMR spectra. 



-] 


..^-.w,™.™*.. 


u8 * , ° i * 


t »■ 




,.^X""' 


i ""■ 


4 .Ij \ 


13 NMR spectrum 
1, 





200 



150 



100 



large -<■ ■ 

low 

-<■ ■ 
(downfield) 

high -<■ ■ 



chemical shift 



field 



frequency 



deshielded 



shielding 



50 5 

>- small 

^high 

(upfield) 

■► low 

>■ shielded 



lactic acid (2-hydroxypropanoic 
acid) 

66.0 (saturated carbon next 
to oxygen) 




19.9 

saturated carbon 176.8 (carbonyl 
not next to oxygen group, C=0) 



NMR spectra were originally 
recorded by varying the applied 
field. They are now recorded by 
variation of the frequency of the 
radio waves and that is done by a 
pulse of radiation. The terms 
'high and low field' are a relic from 
the days of scanning by field 
variation. 



If you are coming back to this chapter 
after reading Chapter 4 you might like 
to know that unsaturated C atoms are 
further deshielded because a rc bond 
has a nodal plane. Jt Bonds have a 
plane with no electron density in at all, 
so electrons in Jt bonds are less 
efficient at shielding the nucleus than 
electrons in it bonds. 



A guided tour of NMR spectra of simple molecules 

We shall first look at NMR spectra of a few simple compounds before looking at unknown struc- 
tures. Our very first compound, hexanedioic acid, has the simple NMR spectrum shown here. The 
first question is: why only three peaks for six carbon atoms? Because of the symmetry of the mole- 
cule, the two carboxylic acids are identical and give one peak at 174.2 p.p.m. By the same token C2 
and C5 are identical while C3 and C4 are identical. These are all in the saturated region 0-50 p.p.m. 
but it is likely that the carbons next to the electron-withdrawing CO2H group are more deshielded 
than the others. So we assign C2/C5 to the peak at 33.2 p.p.m. and C3/C4 to 24.0 p.p.m. 



Why isn't this compound called 
'hexane-l,6-dioic acid'? Well, 
carboxylic acids can only be at the end 
of chains, so no other hexanedioic 
acids are possible: the 1 and 6 are 
redundant. 



62 



3 ■ Determining organic structures 



This spectrum was run in a different 
solvent, DMSO (DiMethyl Sulfoxide); 
hence the brown solvent peaks are in a 
different region and have a different 
form. Wheneveryou first look at a 
spectrum, identify the peaks due to the 
solvent! 




hexanedioic acid 



200 



150 



100 



50 



The peak due to the carbonyl carbon is 
particularly small in this spectrum. This 
is typical for quaternary carbons, i.e. 
carbons that have no protons attached 
to them. 



The bee alarm pheromone (heptan-2-one) has no symmetry so all its seven carbon atoms are differ- 
ent. The carbonyl group is easy to identify (208.8 p. p.m., highlighted in red) but the rest are more diffi- 
cult. Probably the two carbon atoms next to the carbonyl group come at lowest field, while C7 is 
certainly at highest field (13.9 p. p.m.). It is important that there are the right number of signals at about 
the right chemical shift. If that is so, we are not worried if we cannot assign each frequency to a precise 
carbon atom. 




7 6 5 ' 3 ~ 1 

bee alarm pheromone 

heptan-2-one 



200 



150 



100 



50 



The NMR spectrum of BHT tells us that 
the t-butyl groups must also rotate 
rapidly as the three methyl groups give 
only one signal. 



You met BHT on p. 000: its formula is C15H24O and the first surprise in its NMR spectrum is that 
there are only seven signals for the 15 carbon atoms. There is obviously a lot of symmetry; in fact the 
molecule has a plane of symmetry vertically as it is drawn here. The very strong signal at 5 = 30.4 
p. p.m. belongs to the six identical methyl groups on the f-butyl groups and the other two signals in 
the 0-50 p. p.m. range are the methyl group at C4 and the central carbons of the f-butyl groups. In 
the aromatic region there are only four signals as the two halves of the molecule are the same. As with 
the last example, we are not concerned with exactly which is which; we just check that there are the 
right number of signals with the right chemical shifts. 




Me 'BHT' C 15 H 24 
plane of symmetry 



I I 



JJ L 



200 



150 



100 



50 



Paracetamol is a familiar painkiller with a simple structure — it too is a phenol but in addition it 
has an amide on the benzene ring. Its NMR spectrum contains one saturated carbon atom at 24 
p. p.m. (the methyl group of the amide side chain), one carbonyl group at 168 p. p.m., and four other 
peaks at 115, 122, 132, and 153 p. p.m. These are the carbons of the benzene ring. Why four peaks? 
The two sides of the benzene ring are the same because the NHCOCH3 side chain can rotate rapidly 
so that C2 and C6 are the same and C3 and C5 are the same. Why is one of these aromatic peaks in 
the C=0 region at 153 p. p.m.? This must be C4 as it is bonded to oxygen, and it just reminds us that 
carbonyl groups are not the only unsaturated carbon atoms bonded to oxygen (see the chart on p. 
000), though it is not as deshielded as the true C=0 group at 168 p. p.m. 



HO' 




paracetamol 



200 



150 



100 



50 



Nuclear magnetic resonance 



63 



The effects of deshielding within the saturated carbon region 

We have mentioned deshielding several times. The reference compound TMS (Me4Si) has very 
shielded carbon atoms because silicon is more electropositive than carbon. Oxygen moves a saturat- 
ed carbon atom downfield to larger chemical shifts (50-100 p. p.m.) because it is much more elec- 
tro negative than carbon and so pulls electrons away from a carbon atom by polarizing the C-O bond. 
In between these extremes was a CO2H group that moved its adjacent carbon down to around 35 
p. p.m. These variations in chemical shift within each of the 50 p.p.m. regions of the spectrum are a 
helpful guide to structure as the principle is simple. 



Electronegative atoms move adjacent carbon atoms downfield (to larger 8 )by 
deshielding. 



For the carbon atom next to the carboxylic acid, the oxygen atoms are, of course, no longer adjacent 
but one atom further away, so their deshielding effect is not as great. 

The reverse is true too: electropositive atoms move adjacent carbon atoms upheld by shielding. This 
is not so important as there are few atoms found in organic molecules that are more electropositive 
than silicon and so few carbons are more shielded than those in Me 4 Si. About the only important 
elements like this are the metals. When a carbon atom is more shielded than those in TMS, it has a 
negative 8 value. There is nothing odd about this — the zero on the NMR scale is an arbitrary 
point. Table 3.5 shows a selection of chemical shift changes caused to a methyl group by changes in 
electronegativity. 



Table 3.5 


Effect of electro 


negativity on chemical 


shift 








Electronic effect 


Electronegativity of 


Compound 


5(CH 3 ) 


5(CH 3 ) - 


-8.4 






atom bonded to 














carbon 










donation 




1.0 


CH3-U 


-14 


-22.4 




T 




2.2 


CH 3 -H 


-2.3 


-10.7 




weak 




1.8 


CH 3 -SiMe 3 


0.0 


-8.4 





no effect 



2.5 



CH 3 -CH 3 



8.4 







wea 


k 


3.1 


i 




— 


i 




3.5 


Witt 


drawal 


4.1 



CH 3 -NH 2 
CH 3 -C0R 
CH 3 -0H 
CH 3 -F 



26.9 
-30 
50.2 
75.2 



18.5 
-22 
41.8 



The last column in Table 3.5 shows the effect that each substituent has when compared to ethane. 
In ethane there is no electronic effect because the substituent is another methyl group so this column 
gives an idea of the true shift caused by a substituent. These shifts are roughly additive. Look back at 
the spectrum of lactic acid on p. 000: the saturated carbons occur at 19.9 and 66.0. The one at 66.0 is 
next both to an oxygen atom and a carbonyl group so that the combined effect would be about 42 + 
22 = 64 — not a bad estimate. 

NMR is a powerful tool for solving unknown structures 

Simple compounds can be quickly distinguished by NMR. These three alcohols of formula C^jqO 
have quite different NMR spectra. 



We shall look at similar but more 
detailed correlations in Chapters 11 
and 15. 



64 



3 ■ Determining organic structures 





isobutanol 
2-methyl propan-1-ol 


t-butanol 
2-methyl propan-2-ol 










n-butanol 
butan-1-ol 




n-butanol 


isobutanol 


t-butanol 


^^^^OH 


"Y^ oh 


^N)H 


o 
o 


62.9 
36.0 


70.2 
32.0 


69.3 
32.7 


^O^OH 


t~ 


O^OH 


o 

o 


20.3 
15.2 


20.4 


— 



The C atoms have been arbitrarily 
colour-coded. 



The meanings of n-, iso-, and t-were 
covered in Chapter 2 (p. 000). 



,- planes of symmetry 



t. 






H 6 




An epoxide is a three-membered cyclic 
ether. 



Each alcohol has a saturated carbon atom next to oxygen, all close together. Then there are car- 
bons next door but one to oxygen: they are back in the 0-50 p.p.m. region but at its low field end — 
about 30-35 p.p.m.. Notice the similarity of these chemical shifts to those of carbons next to a 
carbonyl group (Table 3.5 on p. 000). In each case we have C-C-O and the effects are about the 
same. Two of the alcohols have carbon(s) one further away still at yet smaller chemical shift (further 
upfield, more shielded) at about 20 p.p.m., but only the n-butanol has a more remote carbon still at 
15.2. The number and the chemical shift of the signals identify the molecules very clearly. 

A more realistic example would be an unknown molecule of formula C3H6O. There are seven 
reasonable structures, as shown. Simple symmetry can distinguish structures A, C, and E from the 
rest as these three have only two types of carbon atom. A more detailed inspection of the spectra 
makes identification easy. The two carbonyl compounds, D and E, each have one peak in the 
150-220 p.p.m. region but D has two different saturated carbon atoms while E has only one. The two 
alkenes, F and G, both have one saturated carbon atom next to oxygen, but F has two normal unsat- 
urated carbon atoms (100-150 p.p.m.) while the enol ether, G, has one normal alkene and one 
unsaturated carbon joined to oxygen. The three saturated compounds (A-C) present the greatest 
problem. The epoxide, B, has two different carbon atoms next to oxygen (50-100 p.p.m.) and one 
normal saturated carbon atom. The remaining two both have one signal in the 0-50 and one in the 
50-100 p.p.m. regions. Only proton NMR (Chapter 11) and, to a certain extent, infrared spec- 
troscopy (which we will move on to shortly) will distinguish them reliably. 

Here are NMR spectra of three of these molecules. Before looking at the solutions, cover up the 
rest of the page and see if you can assign them to the structures above. Try also to suggest which sig- 
nals belong to which carbon atoms. 



spectrum 1 



200 


150 


100 




5( 


) 





spectrum 2 






1 









200 



150 



100 



50 



spectrum 3 



200 



150 



100 



50 



Infrared spectra 



65 



These shouldn't give you too much trouble. The only carbonyl compound with two identical car- 
bons is acetone, Me2CO (E) so spectrum 3 must be that one. Notice the very low field C=0 signal 
(206.6 p.p.m.) typical of a simple ketone. Spectrum 1 has two unsaturated carbons and a saturated 
carbon next to oxygen so it must be F or G. In fact it has to be F as both unsaturated carbons are sim- 
ilar (137 and 116 p.p.m.) and neither is next to oxygen (>150 p.p.m., cf. 206.6 in spectrum 3). This 
leaves spectrum 2, which appears to have no carbon atoms next to oxygen as all chemical shifts are 
less than 50 p.p.m. No compound fits that description (impossible for C3H 6 anyway!) and the two 
signals at 48.0 and 48.2 p.p.m. are suspiciously close to the borderline. They are, of course, next to 
oxygen and this is compound B. 



Infrared spectra 

Functional groups are identified by infrared spectra 

Some functional groups, for example, C=0 or C=C, can be seen in the NMR spectrum because they 
contain carbon atoms, while the presence of others like OH can be inferred from the chemical shifts 
of the carbon atoms they are joined to. Others cannot be seen at all. These might include NH2 and 
NO2, as well as variations around a carbonyl group such as COC1, CO2H, and CONH2. Infrared (IR) 
spectroscopy provides a way of finding these functional groups because it detects the stretching and 
bending of bonds rather than any property of the atoms themselves. It is particularly good at detect- 
ing the stretching of unsymmetrical bonds of the kind found in functional groups such as OH, C=0, 
NH 2 , and N0 2 . 

NMR requires electromagnetic waves in the radio-wave region of the spectrum to make nuclei 
flip from one state to another. The amount of energy needed for stretching and bending individual 
bonds, while still very small, corresponds to rather shorter wavelengths. These wavelengths lie in the 
infrared, that is, heat radiation just to the long wavelength side of visible light. When the carbon 
skeleton of a molecule vibrates, all the bonds stretch and relax in combination and these absorptions 
are unhelpful. However some bonds stretch essentially independently of the rest of the molecule. 
This occurs if the bond is either: 

• much stronger or weaker than others nearby, or 

• between atoms that are much heavier or lighter than their neighbours 

Indeed, the relationship between the frequency of the bond vibration, the mass of the atoms, and the 
strength of the bond is essentially the same as Hooke's law for a simple harmonic oscillator. 



V = 



2kc 



G 



The equation shows that the frequency of the vibration V is proportional to the (root of) a force 
constant/ — more or less the bond strength — and inversely proportional to the (root of) a reduced 
mass fl, that is, the product of the masses of the two atoms forming the bond divided by their sum. 

fi = ^~ 

m 1 +m 2 

Stronger bonds vibrate faster and so do lighter atoms. You may at first think that stronger bonds 
ought to vibrate more slowly, but a moment's reflection will convince you of the truth: which 
stretches and contracts faster, a tight steel spring or a slack steel spring? 

Infrared spectra are simple absorption spectra. The sample is exposed to infrared radiation and 
the wavelength scanned across the spectrum. Whenever energy corresponding to a specific wave- 
length is absorbed, the intensity of the radiation reaching a detector momentarily decreases, and this 
is recorded in the spectrum. Infrared spectra are usually recorded using a frequency measurement 
called wavenumber (cm - ) which is the inverse of the true wavelength A, in centimetres to give con- 
venient numbers (500-4000 cnT ). Higher numbers are to the left of the spectrum because it is real- 
ly wavelength that is being scanned. 



bond vibration in the infrared 



m 1 



contracting 



n\\ 



stretching 



o 



m 1 



m 1 



Hooke's law describes the movement 
of two masses attached to a spring. 
You may have met it if you have studied 
physics. You need not be concerned 
herewith its derivation, just the result. 



66 



3 ■ Determining organic structures 



We need to use another equation here: 



h is Planck's constant and cthe 
velocity of light. 



You should always check the way 
the spectrum was run before 
making any deductions! 



E = hv = h— since X = — 

X v 

The energy, E, required to excite a bond vibration can be expressed as the inverse of a wavelength 
A, or as a frequency V. Wavelength and frequency are just two ways of measuring the same thing. 
More energy is needed to stretch a strong bond and you can see from this equation that larger E 
means higher wavenumbers (cm~ ) or smaller wavelength (cm). 

To run the spectrum, the sample is either dissolved in a solvent such as CHCI3 (chloroform) that 
has few IR absorptions, pressed into a transparent disc with powdered solid KBr, or ground into an 
oily slurry called a mull with a hydrocarbon oil called 'NujoF. Solutions in CHCI3 cannot be used for 
looking at the regions of C-Cl bond stretching nor can Nujol mulls be used for the region of C-H 
stretching. Neither of these is a great disadvantage, especially as nearly all organic compounds have 
some C-H bonds anyway. 




We shall now examine the relationship between bond stretching and frequency in more detail. 
Hooke's law told us to expect frequency to depend on both mass and bond strengths, and we can 
illustrate this double dependence with a series of bonds of various elements to carbon. 

Values chiefly affected by mass of atoms: (lighter atom, higher frequency) 

C-H C-D C-0 C-CI 

3000 cm" 1 2200 cm" 1 1100 cm" 1 700 cm" 1 

Values chiefly affected by bond strength (stronger bond, higher frequency) 



C=0 

2143 cm" 1 



C=0 

1715cm" 1 



C-0 

1100 cm" 1 



Just because they were first recorded in this way, infrared spectra have the baseline at the top and 
peaks going downwards. You might say that they are plotted upside down and back to front. At least 
you are now accustomed to the horizontal scale running backwards as that happens in NMR spectra 
too. A new feature is the change in scale at 2000 cnT so that the right-hand half of the spectrum is 
more detailed than the left-hand half. A typical spectrum looks like this. 



Infrared spectra 



67 



100% 



80% 



60% 




40% 



20% 



4000 3500 3000 2500 2000 1500 

frequency / cm -1 



1000 



500 



There are four important regions of the infrared spectrum 

You will see at once that the infrared spectrum contains many lines, particularly at the right-hand 
(lower frequency) end; hence the larger scale at this end. Many of these lines result from several 
bonds vibrating together and it is actually the left-hand half of the spectrum that is more useful. 

The first region, from about 4000 to about 2500 cm~ is the region for C-H, N-H, and O-H bond 
stretching. Most of the atoms in an organic molecule (C, N, O, for example) are about the same 
weight. Hydrogen is an order of magnitude lighter than any of these and so it dominates the stretch- 
ing frequency by the large effect it has on the reduced mass. The reduced mass of a C-C bond is ( 1 2 x 
12)/(12 + 12), i.e. 144/24 = 6.0. If we change one of these atoms for H, the reduced mass changes to 
(12 x 1)/ (12 + 1), i.e. 12/13 = 0.92, but, ifwe change it instead for F, the reduced mass changes to (12 
x 19)/ (12 + 19), i.e. 228/31 = 7.35. There is a small change when we increase the mass to 19 (F), but 
an enormous change when we decrease it to 1 (H). 

Even the strongest bonds — triple bonds such as feC or C^N — absorb at slightly lower frequencies 
than bonds to hydrogen: these are in the next region from about 2500 to 2000 cirT . This and the other 
two regions of the spectrum follow in logical order of bond strength as the reduced masses are all about 
the same: double bonds such as C=C and C=0 from about 1900-1500 cirT and single bonds at the right- 
hand end of the spectrum. These regions are summarized in this chart, which you should memorize. 

• The regions of the infrared spectrum 

-* increasing energy required to vibrate bond 

■* frequency scale in wavenumbers (cm 1 ) 



4000 



3000 



2000 



1500 



1000 



bonds to 


triple 


double 


single 




hydrogen 


bonds 


bonds 


bonds 




O—H 


C=C 


C=C 


C— O 




N — H 






C — F 




C— H 


C=N 


c=o 


C— CI 





note change 



h 



scale 



Looking back at the typical spectrum, we see peaks in the X-H region at about 2950 cm which are 
the C-H stretches of the CH3 and CH2 groups. The one rather weak peak in the triple bond region 



(2270 cm" ) is of course the C=N group and the strong peak at about 1670 cm~ belongs to the C=0 
group. We shall explain soon why some IR peaks are stronger than others. The rest of the spectrum is 
in the single bond region. This region is not normally interpreted in detail but is characteristic of the 
compound as a whole rather in the way that a fingerprint is characteristic of an individual human 



IR spectra are plotted 'upside 
down' because they record 
transmission (the amount of light 
reaching the detector) rather than 
absorbance. 




^ 



H 



H H 



cyanoacetamide 
(spectrum taken as a Nujol mull 



The concept of reduced mass was 
introduced on p. 000. 



Remember. Hooke's law says that 
frequency depends on both mass and a 
force constant (bond strength). 



68 



3 ■ Determining organic structures 



being — and, similarly, it cannot be 'interpreted'. It is indeed called the fingerprint region. The useful 
information from this spectrum is the presence of the CN and C=0 groups and the exact position of 
the C=0 absorption. 

The X-H region distinguishes C-H, N-H, and O-H bonds 

The reduced masses of the C-H, N-H, and O-H combinations are all about the same. Any difference 
between the positions of the IR bands of these bonds must then be due to bond strength. In practice, 
C-H stretches occur at around 3000 cm" (though they are of little use as virtually all organic com- 
pounds have C-H bonds), N-H stretches occur at about 3300 cm" , and O-H stretches higher still. 
We can immediately deduce that the O-H bond is stronger than N-H which is stronger than C-H. 
IR is a good way to measure such bond strengths. 



Table 3.6 IR bands for bonds to hydrogen 



This may surprise you: you may be used 
to thinking of O-H as more reactive 
than CH. This is, of course, true but, as 
you will see in Chapter 5, factors other 
than bond strength control reactivity. 
Bond strengths will be much more 
important when we discuss radical 
reactions in Chapter 39. 



Bond 



C-H 



Reduced mass, u 



:0.92 



12/13= 
N-H 14/15 = 0.93 

O-H 16/17 = 0.94 

a When not hydrogen-bonded: see below. 



IR frequency, cm 

2900-3200 
3300-3400 
3500-3600 3 



Bond strength, kJ mol 1 

CH 4 : 440 
NH 3 :450 
H 2 0: 500 



The X-H IR stretches are very different in these four compounds. 




4000 3500 3000 2500 2000 1500 
frequency / cm- 1 



4000 3500 3000 2500 2000 1500 
frequency / cm- 3 



1000 



500 




4000 3500 3000 2500 2000 1500 
frequency / cm- 1 



1000 



500 



4000 3500 3000 2500 2000 1500 
frequency / cm- 3 



antisymmetric 
NH 2 stretch 



H ^H 



about 
3400 cm" 1 



symmetric 
NH 2 stretch 



N 
H' "H 



about 
3300 cm" 1 



The IR peak of an NH group is different from that of an NH2 group. A group gives an indepen- 
dent vibration only if both bond strength and reduced mass are different from those of neighbouring 
bonds. In the case of N-H, this is likely to be true and we usually get a sharp peak at about 3300 cm" , 
whether the NH group is part of a simple amine (R2NH) or an amide (RCONHR). The NH2 group is 
also independent of the rest of the molecule, but the two NH bonds inside the NH2 group have iden- 
tical force constants and reduced masses and so vibrate as a single unit. Two equally strong bands 
appear, one for the two N-H bonds vibrating in phase (symmetric) and one for the two N-H bonds 
vibrating in opposition (antisymmetric). The antisymmetric vibration requires more energy and is at 
slightly higher frequency. 



Infrared spectra 



69 



The O-H bands occur at higher frequency, sometimes as a sharp absorption at about 3600 cnT . 
More often, you will see a broad absorption at anywhere from 3500 to 2900 cm This is because OH 
groups form strong hydrogen bonds that vary in length and strength. The sharp absorption at 3600 
cm is the non-hydrogen-bonded OH and the lower the absorption the stronger the H bond. 

Alcohols form hydrogen bonds between the hydroxyl oxygen of one molecule and the hydroxyl 
hydrogen of another. These bonds are variable in length (though they are usually rather longer than 
normal covalent O-H bonds) and they slightly weaken the true covalent O-H bonds by varying 
amounts. When a bond varies in length and strength it will have a range of stretching frequencies 
distributed about a mean value. Alcohols typically give a rounded absorption at about 3300 cm" 
(contrast the sharp N-H stretch in the 



R 

I 



I 



rU 



0" 

: I 

R 

hydrogen bonding in an alcohol 



// 



0— H- 



// 
-0 



the hydrogen-bonded dimer 
of a carboxylic acid 



same region). Carboxylic acids (RCO2H) 
form hydrogen-bonded dimers with two 
strong H bonds between the carbonyl oxy- 
gen atom of one molecule and the acidic 
hydrogen of the other. These also vary 
considerably in length and strength and 
usually give very broad V-shaped 
absorbances. 

Good examples are paracetamol and BHT. Paracetamol has a typical sharp peak at 3330 cm" for 
the N-H stretch and then a rounded 
absorption for the hydrogen-bonded O-H 
stretch from 3300 down to 3000 cm~ in 
the gap between the N-H and C-H 
stretches. By contrast, BHT has a sharp 
absorption at 3600 cm~ as the two large 
and roughly spherical t-butyl groups pre- 
vent the normal H bond from forming. 
100% 



Me 



■kA v - H 



the hydrogen-bonded OH 
group in paracetamol 





4000 3500 3000 2500 2000 1500 

frequency / cm _ l 



1000 



500 



Hydrogen bonds are weak bonds 
formed from electron-rich atoms 
such as or N to hydrogen atoms 
also attached by 'normal' bonds 
to the same sorts of atoms. In 
this diagram of a hydrogen bond 
between two molecules of water, 
the solid line represents the 
'normal' bond and the green 
dotted line the longer hydrogen 
bond. The hydrogen atom is about 
a third of the way along the 
distance between the two oxygen 
atoms. 

H 

hydrogen bond 

■s. 



The C NMR spectra of these two 
compounds are on pp. 000 and 000. 



We can use the N-H and O-H 
absorptions to rule out an 
alternative isomeric structure for 
paracetamol: an ester with an 
NH 2 group instead of an amide 
with NH and OH. This structure 
must be wrong as it would give 
two similar sharp peaks at about 
3300 cm -1 instead of the one 
sharp and one broad peak 
actually observed. 

H 



I 



.A. 




alternative and wrong 
structure for paracetamol 




4000 3500 3000 2500 



2000 1500 
frequency / cm - l 



1000 



500 



in BHT H-bonding is prevented 
by large t-butyl groups 



70 



3 ■ Determining organic structures 



You may be confused the first time you see the IR spectrum of a terminal alkyne, R-C=C-H, 
because you will see a strongish sharp peak at around 3300 cirT that looks just like an N-H stretch. 
The displacement of this peak from the usual C-H stretch at about 3000 cm~ cannot be due to a 
change in the reduced mass and must be due to a marked increase in bond strength. The alkyne C-H 
bond is shorter and stronger than alkane C-H bonds. 



100% 



In Chapter 4, you will see that 
carbon uses an sp 3 orbital to 
make a C-H bond in a saturated 
structure but has to use an sp 
orbital for a terminal alkyne C-H. 
This orbital has one-half s 
character instead of one-quarter s 
character. The electrons in an s 
orbital are held closer to the 
carbon's nucleus than in a p 
orbital, so the sp orbital makes 
for a shorter, stronger C-H bond. 



What are the other peaks in this 
spectrum? 




4000 3500 3000 2500 2000 1500 

frequency / cm - ! 



1000 



500 



1 The summary chart shows some typical peak shapes and frequencies for X-H 
bonds in the region 4000-3000 cm -1 . 



4000 



3800 



3600 



3400 



3200 



3000 



C-H 300C 



non H-bonded 
0-H 3600 



V 




H-bonded 0-H N 
3500-3000 



Derealization is covered in 
Chapter 7; forthe moment, just 
accept that both NO bonds are 
the same. 



The double bond region is the most important in IR spectra 

In the double bond region, there are three important absorptions, those of the carbonyl (C=0), 
alkene (C=C), and nitro (NO2) groups. All give rise to sharp bands: C=0 to one strong (i.e. intense) 
band anywhere between 1900 and 1500 cm~ ; C=C to one weak band at about 1640 cm~ ; and NO2 
to two strong (intense) bands in the mid-1500s and mid-1300s cm . The number of bands is easily 
dealt with. Just as with OH and NH 2 , it is a matter of how many identical bonds are present in the 
same functional group. Carbonyl and alkene clearly have one double bond each. The nitro group at 
first appears to contain two different groups, N + -CF and N=0, but derealization means they are 
identical and we see absorption for symmetrical and antisymmetrical stretching vibrations. As with 
NH2, more work is needed for the antisymmetrical vibration which occurs at higher frequency 
(>1500pluscm _1 ). 



Infrared spectra 



71 



derealization in the nitro group 


symmetric 




N0 2 stretch (-1350) 


r can be drawn as R 




(both NO bonds 


R 


^m identical) , |\|. 
0^©^0 s " ^0 


<A> 





(-) 0"©^0 (-) 



antisymmetric 
N0 2 stretch (-1550) 



-) Or 0^*0 (- 



The strength of an IR absorption depends on dipole moment 

Now what about the variation in strength (i.e. intensity, the amount of energy absorbed)? The 
strength of an IR absorption varies with the change of dipole moment when the bond is stretched. If 
the bond is perfectly symmetrical, there is no change in dipole moment and there is no IR absorp- 
tion. Obviously, the C=C bond is less polar than either C=0 or N=0 and is weaker in the IR. Indeed 
it may be absent altogether in a symmetrical alkene. By contrast the carbonyl group is very polar 
(Chapter 4) and stretching it causes a large change in dipole moment and C=0 stretches are usually 
the strongest peaks in the IR spectrum. You may also have noticed that O-H and N-H stretches are 
stronger than C-H stretches (even though most organic molecules have many more C-H bonds 
than O-H or N-H bonds): the reason is the same. 



Dipole moments 

Dipole moment depends on the variation in distribution of 
electrons along the bond, and also its length, which is why 
stretching a bond can change its dipole moment. For 
bonds between unlike atoms, the larger the difference in 
electronegativity, the greater the dipole moment, and the 
more it changes when stretched. For identical atoms 



(C=C, for example) the dipole moment, and its capacity to 
change with stretching, is much smaller. Stretching 
frequencies for symmetrical molecules are measured 
using Raman spectra. This is an IR-based technique using 
scattered light that relies on polarizability of bonds. 
Raman spectra are outside the scope of this book. 



Contrast the term 'strength' 
applied to absorption and to 
bonds. A stronger absorption is a 
more intense absorption — i.e. 
one with a big peak. A strong bond 
on the other hand has a higher 
frequency absorption (other 
things being equal). 



This is a good point to remind you of the various deductions we have made so far about IR spectra. 

• Absorptions in IR spectra 



Position of band depends on - 



reduced mass of atoms 



bond strength 



light atoms give 
high frequency 
strong bonds give 
high frequency 



Strength of band depends on — > change in dipole moment large dipole moment gives 

strong absorption 



Width of band depends on - 



hydrogen bonding 



strong H bond gives 
wide peak 



We have seen three carbonyl compounds so far in this chapter and they all show peaks in the right 



region (around 1700 cmT ) even though one is a carboxylic acid, one a ketone, and one an amide. 
We shall consider the exact positions of the various carbonyl absorptions in Chapter 15 after we have 
discussed some carbonyl chemistry. 




HO 




Y 



hexanedioic acid 
1720 cm" 1 



heptan-2-one 
1710 cm 1 



paracetamol 
1667 cm" 1 



72 



3 ■ Determining organic structures 



Table 3.7 Single bonds 



Pair of atoms Reduced mass Bond strength 



The single bond region is used as a molecular fingerprint 

The region below 1500 crrT is where the single bond vibrations occur. Here our hope that individ- 
ual bonds may vibrate independently of the rest of the molecule is usually doomed to disappoint- 
ment. The atoms C, N, and O all have about the same atomic weight and C-C, C-N, and C-O single 
bonds all have about the same strength. 

In addition, C-C bonds are likely to be joined to 
other C-C bonds with virtually identical strength and 
reduced mass, and they have essentially no dipole 
moments. The only one of these single bonds of any 
value is C-O which is polar enough and different 
enough (Table 3.7) to show up as a strong absorption at 
about 1 100 cnT . Some other single bonds such as C-Cl 
(weak and with a large reduced mass) are quite useful at 
about 700 cnT . Otherwise the single bond region is 
usually crowded with hundreds of absorptions from 
vibrations of all kinds used as a 'fingerprint' characteris- 
tic of the molecule but not really open to interpretation. 

Among the hundreds of peaks in the fingerprint 
region, there are some of a quite different kind. 
Stretching is not the only bond movement that leads to 
IR absorption. Bending of bonds, particularly C-H and 
N-H bonds, also leads to quite strong peaks. These are 
called deformations. Bending a bond is easier than stretching it (which is easier, stretching or bending 
an iron bar?). Consequently, bending absorptions need less energy and come at lower frequencies than 
stretching absorptions for the same bonds. These bands may not often be useful in identifying mole- 
cules, but you will notice them as they are often strong (they are usually stronger than C=C stretches for 
example) and may wonder what they are. 

Finally in this section, we summarize all the useful absorptions in the fingerprint region. Please be 
cautious in applying these as there are other reasons for bands in these positions. 



C-C 


6.0 


350 kJ moP 1 


C-N 


6.5 


305 kJ mol" 1 


C-O 


6.9 


360 kJ mol" 1 


Table 3.8 Useful deformat 


ons 


(bend 


ng vibrations) 




Group 


Frequency, cm" 1 


Strength 


CH 2 


1440-1470 


medium 


CH 3 


-1380 


medium 


NH 2 


1550-1650 


medium 



You may not yet understand all the 
terms in Table 3.9, but you will find it 
useful to refer back to later. 





Table 3.9 Useful absorptions in the fingerprint region 


Frequency, cm" 1 


Strength 


Group 


Comments 


1440-1470 


medium 


CH 2 


deformation (present in nujol) 


-1380 


medium 


CH 3 


deformation (present in nujol) 


-1350 


strong 


N0 2 


symmetrical N=0 stretch 


1250-1300 


strong 


P=0 


double bond stretch 


1310-1350 


strong 


S0 2 


antisymmetrical S=0 stretch 


1120-1160 


strong 


S0 2 


symmetrical S=0 stretch 


-1100 


strong 


C-O 


single bond stretch 


950-1000 


strong 


C=CH 


trans alkene (out-of-plane deformation) 


-690 and -750 


strong 


Ar-H 


five adjacent Ar-H (out-of-plane) 


-750 


strong 


Ar-H 


four adjacent Ar-H (out-of-plane) 


-700 


strong 


C-CI 


single bond stretch 





Mass spectra, NMR, and IR combined make quick 
identification possible 

If these methods are each as powerful as we have seen on their own, how much more effective they 
must be together. We shall finish this chapter with the identification of some simple unknown 



Mass spectra, NMR, and IR combined make quick identification possible 



73 



compounds using all three methods. The first is an industrial emulsifier used to blend solids and liquids 
into smooth pastes. Its electron impact mass spectrum has peaks at 75 and 74 (each about 20%) and a 
base peak at 58. The two peaks at 75 and 74 cannot be isotopes of bromine as the separation is only one 
mass unit, nor can 75 be a C peak as it is far too strong. It looks as though 75 might be the molecular 
ion and 74 an unusual loss of a hydrogen atom. However a chemical ionization mass spectrum reveals a 
molecular ion at 90 (MH + ) and hence the true molecular ion at 89. An odd molecular weight (89) sug- 
gests one nitrogen atom, and high resolution mass spectrometry reveals that the formula is C4H1 jNO. 

the electron impact mass spectrum of an industrial emulsifier 
100- 



80- 



60- 



40- 



20- 



_ 



-. 



OH 



NH, 



_ 



r — t— 

15 30 45 60 75 90 105 

The C NMR spectrum has only three peaks so two carbon atoms must be the same. There is one 
signal for saturated carbon next to oxygen, and two for other saturated carbons, one more downfield 
than the other. The IR spectrum reveals a broad peak for an OH group with two sharp NH 2 peaks 
just protruding. If we put this together, we know we have C-OH and C-NH 2 . Neither of these car- 
bons can be duplicated (as there is only one O and only one N!) so one of the remaining carbons 
must be duplicated. 

13 C NMR spectrum 



jlL 



200 



150 



100 



50 



100% 




4000 



3500 



3000 2500 



2000 1500 

frequency / cm -1 



1000 



500 



74 



3 ■ Determining organic structures 



The next stage is one often overlooked. We don't seem to have much information, but try and put 
the two fragments together, knowing the molecular formula, and there's very little choice. The car- 
bon chain (shown in red) could either be linear or branched and that's it! 



By chain terminating we mean 
only attachable to one other 
atom. 



linear carbon chain 



branched carbon chain 





HO 



NH 2 



H0 O<C 



H 2 N 



NH 2 



OH 



B 

There is no room for double bonds or rings because we need to fit in the eleven hydrogen atoms. 
We cannot put N or O in the chain because we know from the IR that we have the chain terminating 
groups OH and NH2. Of the seven possibilities only the last two, A and B, are serious since they alone 
have two identical carbon atoms (the two methyl groups in each case); all the other structures would 
have four separate signals in the NMR. How can we choose between these? The base peak in the mass 
spectrum was at 58 and this fits well with a fragmentation of one structure but not of the other: the 
wrong structure would give a fragment at 59 and not 58. The industrial emulsifier is 2-amino-2- 
methylpropan-1-ol. 



HO' 



NH 2 



fragmentation 



,CH 2 



HO' 



© 



2-amino-2-methylpropan-l-ol 



CH 3 0" = 31 

not seen because 

not a cation 



NH 3 



C 3 H 8 N + = 58 



H 2 N X ^ S V' 



OH fragmentation 



-^ H 2 N' 



l-amino-2-methylpropan-2-ol 



,CH 2 



CH 4 N - = 30 

not seen because 

not a cation 



© 



OH 



C3H7O + = 59 



Double bond equivalents help in the search for a structure 

The last example was fully saturated but it is usually a help in deducing the structure of an unknown 
compound if, once you know the atomic composition, you immediately work out how much unsat- 
uration there is. This is usually expressed as 'double bond equivalents'. It may seem obvious to you 
that, if C4HHNO has no double bonds, then C4H9NO (losing two hydrogen atoms) must have one 
double bond, C4H7NO two double bonds, and so on. Well, it's not quite as simple as that. Some pos- 
sible structures for these formulae are shown below. 

some structures for C 4 H 9 N0 



NH 2 




NH 3 



some structures for C4H7NO 
,NH 2 




d 




Mass spectra, NMR, and IR combined make quick identification possible 



75 



Some of these structures have the right number of double bonds (C=C and C=0), one has a triple 
bond, and three compounds use rings as an alternative way of 'losing' some hydrogen atoms. Each 
time you make a ring or a double bond, you have to lose two more hydrogen atoms. So double bonds 
(of all kinds) and rings are called Double Bond Equivalents (DBEs). 

You can work out how many DBEs there are in a given atomic composition just by making a 
drawing of one possible structure (all possible structures have the same number of DBEs). 
Alternatively, you can calculate the DBEs if you wish. A saturated hydrocarbon with « carbon atoms 
has (2« + 2) hydrogens. Oxygen doesn't make any difference to this: there are the same number of 
Hs in a saturated ether or alcohol as in a saturated hydrocarbon. 

So, for a compound containing C, H, and O only, take the actual number of hydrogen atoms away 
from (2n + 2) and divide by two. Just to check that it works, for the unsaturated ketone CyH^O the 
calculation becomes: 

1 Maximum number of H atoms for 7 Cs 

2 Subtract the actual number of H atoms (12) 

3 Divide by 2 to give the DBEs 4/2 = 2 C 7 H 12 = two DBE 

The unsaturated ketone does indeed have an alkene and a carbonyl group. The unsaturated cyclic 
acid has: 16 - 10 = 6 divided by 2 = 3 DBEs and it has one alkene, one C=0 and one ring. Correct. 
The aromatic ether has 16-8 = 8 divided by 2 gives 4 DBEs and it has three double bonds in the ring 
and the ring itself. Correct again. 

Nitrogen makes a difference. Every nitrogen adds one extra hydrogen atom because nitrogen can 
make three bonds. This is one fewer hydrogen to subtract. The formula becomes: subtract actual 
number of hydrogens from {In + 2), add one for each nitrogen atom, and divide by two. We can try 
this out too. 



In 


f 2 = 


16 


16- 


- 12 = 


= 4 


4/2 


= 2 






saturated hydrocarbon C 7 H 16 




saturated alcohol C7H16O 
OH 




saturated ether C 7 H 16 




All have (2n + 2) H atoms 



C0 2 H 




C 7 H 10 = two DBE 




OMe 



C7H0O = four DBE 



saturated C 7 compound with nitrogen 
NH 2 



C 7 H 17 N = (2n + 3)Hs 




NMe 




NH 2 



NMe, 




C 7 H 15 N0 2 = one DBE C 7 H 13 N0 = two DBE C 7 H 9 N0= four DBE C 7 H 10 N 2 = four DBE 



The saturated compound has (In + 3) Hs instead of (2» + 2). The saturated nitro compound has 
(2« + 2) = 16 less 15 (the actual number of Hs) plus one (the number of nitrogen atoms) = 2. Divide 
this by 2 and you get 1 DBE, which is the N=0 bond. The last compound (we shall meet this later as 
'DMAP') has: 

1 Maximum number of H atoms for 7 Cs In + 2 = 16 

2 Subtract the actual number of H atoms (10) 16 - 10 = 6 

3 Add number of nitrogens 6 + 2 = 8 

4 Divide by 2 to give the DBEs 8/2 = 4 

There are indeed three double bonds and a ring, making four in all. You would be wise to check that 
you can do these calculations without much trouble. 

If you have other elements too it is simpler just to draw a trial structure and find out how many 
DBEs there are. You may prefer this method for all compounds as it has the advantage of finding one 
possible structure before you really start! One good tip is that if you have few hydrogens relative to 
the number of carbon atoms (and at least four DBEs) then there is probably an aromatic ring in the 
compound. 



Do not confuse this calculation 
with the observation we made 
about mass spectra that the 
molecular weight of a compound 
containing one nitrogen atom 
must be odd. This observation 
and the number of DBEs are, of 
course, related but they are 
different calculations made for 
different purposes. 



76 



3 ■ Determining organic structures 



> Working out the DBEs for an unknown compound 

1 Calculate the expected number of Hs in the saturated structure 

(a) For C n there would be: 2n + 2 Hs if C, H, O only 

(b) For C n N m there would be 2n + 2 + m Hs 

2 Subtract the actual number of Hs and divide by 2. This gives the DBEs 

3 If there are other atoms (CI, B, P, etc.) it is best to draw a trial structure 

4 If there are few Hs, e.g. less than the number of Cs, suspect a benzene ring 

5 A benzene ring has four DBEs (three for the double bonds and one for the ring) 

6 A nitro group has one DBE only 



HBr 



-*- ? 



acrolein 
(propenal) 



ethylene glycol 
(ethane-l,2-diol) 



An unknown compound from a chemical reaction 

Our second example addresses a situation very common in chemistry — working out the structure of 
a product of a reaction. The situation is this: you have treated propenal (acrolein) with HBr in 
ethane- 1,2-diol (or glycol) as solvent for one hour at room temperature. Distillation of the reaction 
mixture gives a colourless liquid, compound X. What is it? 
mass spectrum of compound X 



100 
90- 
80 

70- 
« 60 

CD 

■o 

I 50 

£ 40 

30 

20 

10 





181.0 



108.9 



73.0 



101.0 



80.9 



Mill 



152.0 



136.9 



122.9 



11 flh 1 i-' r i 



^J-L 



163.0 



uai 



100 



150 



13 C NMR spectrum for propenal 



vf*»y ^nfimiHfK »i 0mftd^ &* jmi*ii**i**/f+**Y&ilit4tiftS l *4* */", » f t^ ^mH tjI ^m ■rtufV*' ■ ■»-w^**^ > — V^ ^< w ^-*v^> ■ Www^*»MiiWmi»«v/y ' " "W **** * * wOWWlW^ ff ^ M titw 



200 



150 



100 



50 







13 C NMR spectrum for compound X 



200 



150 



100 



50 



Mass spectra, NMR, and IR combined make quick identification possible 



77 



100% 




4000 3500 3000 2500 2000 1500 

frequency / cm -1 



1000 



500 



The mass spectrum shows a molecular ion (181) much heavier than that of the starting material, 
C3H4O = 56. Indeed it shows two molecular ions at 181/179 typical of a bromo-compound, so it 
looks as if HBr has added to the aldehyde somehow. High resolution reveals a formula of CsHgBrC^ 
and the five carbon atoms make it look as though the glycol has added in too. If we add everything 
together we find that the unknown compound is the result of the three reagents added together less 
one molecule of water. A trial structure reveals one DBE. 



HBr 



X 
C 5 HgBr0 2 



C a H 4 



C2H602 



HBr 



C5HnBr03 



C 5 H 9 Br0 2 + H 2 



The next thing is to see what remains of the propenal. The NMR spectrum of CH 2 =CH-CHO 
clearly shows one carbonyl group and two carbons on a double bond. These have all disappeared in 
the product and for the five carbon atoms we are left with four signals, two saturated, one next to 
oxygen, and one at 102.6 p.p.m. just creeping into the double bond region. It can't be an alkene as an 
alkene is impossible with only one carbon atom! The IR spectrum gives us another puzzle — there 
appear to be no functional groups at all! No OH, no carbonyl, no alkene — what else can we have? 
The answer is an ether — or rather two ethers as there are two oxygen atoms. Now that we suspect an 
ether, we can look for the C-O single bond stretch in the IR spectrum and find it at 1 128 cm - . Each 
ether oxygen must have a carbon atom on each side of it. Two of these could be the same, but where 
are the rest? 

We can solve this problem with a principle you may have guessed at before. If one oxygen atom 
takes a saturated carbon atom downfield to 50 p.p.m. or more, what could take a carbon downfield 
to 100 p.p.m. or more? We have established that chemical shifts are roughly additive so two oxygen 
atoms would just do. This would give us a fragment C-O-C-O-C accounting for three of the five 
carbon atoms. If you try and join the rest up with this fragment, you will find that you can't do it 
without a double bond, for example, the structure in the margin. 

But we know we haven't got a double bond, (no alkene and no C=0) so the DBE must be a ring. 
You might feel uncomfortable with rings, but you must get used to them. Five-, six-, and seven-mem- 
bered rings are very common. In fact, most known organic compounds have rings in them. We could 
join the skeleton of the present molecule up in many rings of various sizes like this one in the margin. 

But this won't do as it would have five different carbon atoms. It is much more likely that the basic 
skeletons of the organic reagents are preserved, that is, that we have a two-carbon and a three-carbon 
fragment joined through oxygen atoms. This gives four possibilities. 




Br 



Br 




Br 





no double bond 
too many Hs 




Me 



C5n^-^Br02 



one double bond 
right amount of Hs 



CH^O 




C 5 H 9 Br0 2 



Br 




OMe 



78 



3 ■ Determining organic structures 



These are all quite reasonable, though we might prefer the third as it is easier to see how it derives 
from the reagents. A decision can easily be reached from the base peak in the mass spectrum at 73. 
This is a fragment corresponding to the five-membered ring and not to the six-membered ring. The 
product is in fact the third possibility. 



Br 



© 
- 




fragmentation 



not seen- 

uncharged 

radical 




73 



HBr 

V 



HO 



~V~ 



Br s , 




Problems 



Looking forward to Chapters 1 1 and 14 

We have only begun to explore the intricate world of identification of structure by spectroscopy. It is 
important that you recognize that structures are assigned, not because of some theoretical reason or 
because a reaction 'ought' to give a certain product, but because of sound evidence from spectra. You 
have seen three powerful methods — mass spectra, C NMR, and IR spectroscopy in this chapter. In 
Chapter 11 we introduce the most important of all — proton ( H) NMR and, finally, in Chapter 14 
we shall take each of these a little further and show how the structures of more complex unknown 
compounds are really deduced. The last problem we have discussed here is not really solvable with- 
out proton NMR and in reality no-one would tackle any structure problem without this most pow- 
erful of all techniques. From now on spectroscopic evidence will appear in virtually every chapter. 
Even if we do not say so explicitly every time a new compound appears, the structure of this com- 
pound will in fact have been determined spectroscopically. Chemists make new compounds, and 
every time they do they characterize the compound with a full set of spectra. No scientific journal 
will accept that a new compound has been made unless a full description of all of these spectra are 
submitted with the report. Spectroscopy lets the science of organic chemistry advance. 



1. How does the mass spectrum give evidence of isotopes in the 
compounds of bromine, chlorine, and carbon. Assuming the 
molecular ion of each of these compounds is of 100% abundance, 
what peaks (and in what intensity) would appear around that 
mass number? (a) C2H5BrO, (b) Cgo> (c) Cgl^BrCl? Give in 
cases (a) and (c) a possible structure for the compound. What 
compound is (b)? 

2. The 13 C NMR spectrum for ethyl 
benzoate contains these peaks: 17.3, 
61.1, 100-150 p. p.m. (four peaks), and 
166.8 p.p.m. Which peak belongs to 
which carbon atom? 




ethyl benzoate 



3. The thinner used in typists' correction fluids is a single com- 
pound, C2H3CI3, having 13 C NMR peaks at 45.1 and 95.0 p.p.m. 
What is its structure? A commercial paint thinner gives two spots 
on thin layer chromatography and has C NMR peaks at 7.0, 
27.5, 35.2, 45.3, 95.6, and 206.3 p.p.m. Suggest what compounds 
might be used to make up this thinner. 

4. The 'normal' O-H stretch (i.e. without hydrogen bonding) 
comes at about 3600 cm~ . What is the reduced mass (|l) for 



O-H? What happens to the reduced mass when you double the 
atomic weight of each atom in turn, that is, what is |X for O-D and 
what is |X for S-H? In fact, both O-D and S-H stretches come at 
about 2500 cm -1 . 

5. Four compounds, each having the formula C3H5NO, have the 
IR spectra summarized here. What are their structures? Without 

C NMR data, it may be easier to tackle this problem by first 
writing down all the possible structures for C3H5NO. In what 
specific ways would C NMR data help? 

(a) One sharp band above 3000 cm~ ; one strong band at about 
1700 cirT 1 

(b) Two sharp bands above 3000 cm - ; two bands between 1600 
and 1700 cm -1 

(c) One strong broad band above 3000 cm ; a band at about 
2200 cm" 1 

6. Four compounds having the molecular formula C4H6O2 have 
the IR and C NMR spectra given below. How many DBEs are 
there in C4H6O2? What are the structures of the four com- 
pounds? You might again find it helpful to draw out some or all 
possibilities before you start. 



Problems 



79 



(a) IR: 1745 cirT 1 ; 13 C NMR: 214, 82, 58, and 41 p.p.m. 

(b) IR: 3300 (broad) cm" 1 ; 13 C NMR: 62 and 79 p.p.m. 

(c) IR: 1770 cirT 1 ; I3 C NMR: 178, 86, 40, and 27 p.p.m. 

(d) IR: 1720 and 1650 (strong) cm" 1 ; 13 C NMR: 165, 131, 133, 
and 54 p.p.m. 

7. Three compounds of molecular formula C4HgO have the IR 
and C NMR spectra given below. Suggest a structure for each 
compound, explaining how you make your deductions, 
compound A IR: 1730 cm" 1 ; 13 C NMR: 13.3, 15.7, 45.7, and 

201.6 p.p.m. 

compound B IR: 3200 (broad) cm" 1 ; 13 C NMR: 36.9, 61.3, 
117.2, and 134.7 p.p.m. 

compound C IR: no peaks except CH and fingerprint; C NMR: 
25.8 and 67.9 p.p.m. 

compound D IR: 3200 (broad) cm" 1 ; 13 C NMR: 15.2, 20.3, 36.0, 
and 62.9 p.p.m. 

Compound A reacts with NaBH4 to give compound D. 
Compound B reacts with hydrogen gas over a palladium catalyst 
to give the same compound D. Compound C reacts with neither 
reagent. Suggest a structure for compound D from the data given 
and explain the reactions. (Note. H2 reduces alkenes to alkanes in 
the presence of a palladium catalyst.) 

8. You have dissolved f-BuOH (Me 3 COH) in MeCN with an 
acid catalyst, left the solution overnight, and found crystals with 
the following characteristics there in the morning. What are they? 



IR: 3435 and 1686 cm" 1 

13 C NMR: 169, 50, 29, and 25 p.p.m. 



H + 



*\ MeCN 



*~ ? 



mass spectrum (%): 115 (7), 100 (10), 64 (5), 60 (21), 59 (17), 58 
(100), and 56 (7). (Don't try to assign all of these!) 

9. How many isomers of trichlorobenzene 
are there? The 1,2,3-trichloro isomer is 
illustrated. Could they be distinguished by 
13 C NMR? 

10. How many signals would you expect in the C NMR of the 
following compounds? 







r 



B 
C0 2 H 



H0 2 C 



„C0 2 H 



H0 2 C 



IN 



OH 



11. How would mass spectra help you distinguish these structures? 




Structure of molecules 



4 



Connections 


Building on: 




Arriving at: 


Looking forward to: 


• How organic structures are drawn ch2 


• 


How we know that electrons have 


• Mechanisms depend on molecular 


• Evidence used to determine organic 




different energies 


orbitals ch5 


structure ch3 


• 


How electrons fit into atomic orbitals 


• Conjugation ch7 




• 


How atomic orbitals combine to make 


• 1 H NMR involves molecular orbitals 






molecular orbitals 


chll 




• 


Why organic molecules have linear, 


• Reactivity derives from energies of 






planar, or tetrahedral structures 


molecular orbitals ch3 




• 


Connection between shape and 
electronic structure 






• 


A true system of molecular orbital 
energies for simple molecules 






• 


Why such rigour is not possible for 
typical organic molecules 






• 


Predicting the locations of lone pairs 
and empty orbitals 






• 


Interaction between theory and 
experiment 





Note from the authors to all readers 

This chapter contains mathematical material that some readers may find daunting. Organic chem- 
istry students come from many different backgrounds since organic chemistry occupies a middle 
ground between the physical and the biological sciences. We hope that those from a more physical 
background will enjoy the material as it is. If you are one of those, you should work your way 
through the entire chapter. If you come from a more biological background, especially if you have 
done little maths at school, you may lose the essence of the chapter in a struggle to understand the 
equations. We have therefore picked out the more mathematical parts in boxes and you should 
abandon these parts (and any others!) if you find them too alien. The general principles behind the 
chapter — why molecules have the structures they do — are obviously so important that we cannot 
omit this essential material but you should try to grasp the principles without worrying too much 
about the equations. The ideas of atomic orbitals overlapping to form bonds, the molecular orbitals 
that result, and the shapes that these orbitals impose on organic molecules are at least as central for 
biochemistry as they are for organic chemistry. Please do not be discouraged but enjoy the challenge. 

Introduction 




You may recognize the model above as DNA, the molecule that carries the genetic information for all 
life on earth. It is the exact structure of this compound that determines precisely what a living thing 



82 



4 ■ Structure of molecules 



is — be it man or woman, frog, or tree — and even more subtle characteristics such as what colour eyes 
or hair people have. 

What about this model? 

You may also have recognized this molecule as 
buckminsterfullerene, a form of carbon that received 
enormous interest in the 1980s and 1990s. The 
question is, how did you recognize these two com- 
pounds? You recognized their shapes. All molecules 
are simply groups of atoms held together by electrons 
to give a definite three-dimensional shape. What 
exactly a compound might be is determined not only 
by the atoms it contains, but also by the arrange- 
ment of these atoms in space — the shape of the 
molecule. Both graphite and buckminsterfullerene 
are composed of carbon atoms only and yet their 
properties, both chemical and physical, are completely 
different. 

There are many methods available to chemists 
and physicists to find out the shapes of molecules. 
One of the most recent techniques is called 
Scanning Tunnelling Microscopy (STM), which is 
the closest we can get to actually 'seeing' the atoms 
themselves. 

Most techniques, for example, X-ray or elect- 
ron diffraction, reveal the shapes of molecules indi- 
rectly. 

In Chapter 3 you met some of the spec- 
troscopic methods frequently used by 
organic chemists to determine the shape of 
molecules. Spectroscopy would reveal the 
structure of methane, for example, as 
tetradral — the carbonatom in the centre of 
a regular tetra-hedron with the hydrogen 
atoms at the corners. In this chapter we are 
going to discuss why compounds adopt the 
shapes that they do. 

This tetrahedral structure seems to be 
very important — other molecules, both 
organic and inorganic, are made up of many 
tetrahedral units. What is the origin of this 
tetrahedral structure? It could simply arise 
from four pairs of electrons repelling each 
other to get as far as possible from each 
other. That would give a tetrahedron. 




The dark brown blobs in this STM 
picture recorded at a temperature 
of 4 K are individual oxygen atoms 
adsorbed on a silver surface. The 
light blobs are individual ethylene 
(ethene) molecules. Ethylene will 
only adsorb on silver if adjacent to 
an oxygen atom. This is an atomic 
scale view of a very important 
industrial process — the 
production of ethylene oxide from 
ethylene and oxygene using a 
silver catalyst. 



The picture on the right is an X-ray 
structure of a catenane — a 
molecule consisting of two 
interlocking rings joined like two 
links in a chain. The key to the 
synthesis depends on the self- 
stacking of the planar structures 
priorto ring closure. 





H 



methane is tetrahedral 



the H atoms form 
a tetrahedron 



methane is 
tetrahedral 



Atomic structure 



83 






tetrahedral methane 
four bonds and no lone pairs 



This simple method of deducing the structure of molecules is called Valence Shell Electron Pair 
Repulsion Theory (VSEPRT). It says that all electron pairs, both bonding and nonbonding, in the 
outer or valence shell of an atom repel each other. This simple approach predicts (more or less) the 
correct structures for methane, ammonia, and water with four electron pairs arranged tetrahedrally 
in each case. 

VSEPRT seems to work for simple structures but surely there must be more to it than this? Indeed 
there is. If we really want to understand why molecules adopt the shapes they do, we must look at the 
atoms that make up the molecules and how they combine. By the end of this chapter, you should be 
able to predict or at least understand the shapes of simple molecules. For example, why are the bond 
angles in ammonia 107°, while in hydrides of the other elements in the same group as nitrogen, PH 3 , 
AsH 3 , and SbH 3 , they are all around 90°? Simple VSEPRT would suggest tetrahedral arrangements 
for each. 



d 



H > H 



tetrahedral ammonia 
three bonds and one lone pair 



Atomic structure 

You know already what makes up an atom — protons, neutrons, and electrons. The protons and 
neutrons make up the central core of an atom — the nucleus — while the electrons form some 
sort of cloud around it. As chemists, we are concerned with the electrons in atoms and more impor- 
tantly with the electrons in molecules: chemists need to know how many electrons there are in a sys- 
tem, where they are, and what energy they have. Before we can understand the behaviour of electrons 
in molecules, we need to look closely at the electronic structure of an atom. Evidence first, theory 
later. 

Atomic emission spectra 

Many towns and streets are lit at night by sodium vapour lamps. You will be familiar with their warm 
yellow- orange glow but have you ever wondered what makes this light orange and not white? The 
normal light bulbs you use at home have a tungsten filament that is heated white hot. You know that 
this white light could be split by a prism to reveal the whole spectrum of visible light and that each 
of the different colours has a different frequency that corresponds to a distinct energy. But where 
does the orange street light come from? If we put a coloured filter in front of our white light, it 
would absorb some colours of the spectrum and let other colours through. We could make orange 
light this way but that is not how the street lights work — they actually generate orange light and 
orange light only. Inside these lights is sodium metal. When the light is switched on, the sodium 
metal is slowly vaporized and, as an electric current is passed through the sodium vapour, an orange 
light is emitted. This is the same colour as the light you get when you do a flame test using a sodium 
compound. 

The point is that only one colour light comes from a sodium lamp and this must have one specif- 
ic frequency and therefore one energy. It doesn't matter what energy source is used to generate the 
light, whether it be electricity or a Bunsen burner flame; in each case light of one specific energy is 
given out. Looking at the orange sodium light through a prism, we see a series of very sharp lines 
with two particularly bright orange lines at around 600 nm. Other elements produce similar spec- 
tra — indeed two elements, rubidium and cesium, were discovered by Robert Bunsen after studying 
such spectra. They are actually named after the presence of a pair of bright coloured lines in their 
spectra — cesium from the Latin caesius meaning bluish grey and rubidium from the Latin rubidus 
meaning red. Even hydrogen can be made to produce an atomic spectrum and, since a hydrogen 
atom is the simplest atom of all, we shall look at the atomic spectrum of hydrogen first. 

If enough energy is supplied to a hydrogen atom, or any other energy © © 

atom, an electron is eventually knocked completely out of the atom. ^ *~ " + e 

In the case of hydrogen a single proton is left. This is, of course, the h atom proton electron 

ionization of hydrogen. 

What if we don't quite give the atom enough energy to remove an electron completely? It's 
not too hard to imagine that, if the energy is not enough to ionize the atom, the electron would be 



(?X» 



tetrahedral water 
two bonds and two lone pairs 



Quantum mechanics tells us that 
energy is quantized. Light does 
not come in a continuous range of 
energies but is divided up into 
minute discrete packets (quanta) 
of different noncontinuous 
(discrete) energies. The energy of 
each of these packets is related 
to the frequency of the light by a 
simple equation: E= r7v(Eisthe 
energy, vthe frequency of the 
light, and h is Planck's constant). 
The packet of light released from 
sodium atoms has the frequency 
of orange I ight and the 
corresponding energy. 



84 4 ■ Structure of molecules 



'loosened' in some way — the atom absorbs this energy and the electron moves further away from the 
nucleus and now needs less energy to remove it completely. The atom is said to be in an excited state. 
This process is a bit like a weight lifter lifting a heavy weight — he can hold it above his head with 
straight arms (the excited state) but sooner or later he will drop it and the weight will fall to the 
ground. This is what happens in our excited atom — the electron will fall to its lowest energy, its 
ground state, and the energy put in will come out again. This is the origin of the lines in the atomic 
spectra not only for hydrogen but for all the elements. The flame or the electric discharge provides 
the energy to promote an electron to a higher energy level and, when this electron returns to its 
ground state, this energy is released in the form of light. 

Line spectra are composed of many lines of different frequencies, which can only mean that there 
must be lots of different energy transitions possible, but not just any energy transitions. Quantum 
mechanics says that an electron, like light, cannot have a continuous range of energies, only certain 
definite energies, which in turn means that only certain energy transitions are possible. This is rather 
like trying to climb a flight of stairs — you can jump up one, two, five, or even all the steps if you have 
enough energy but you cannot climb up half or two-thirds of a step. Likewise coming down, you can 
jump from one step to any other — lots of different combinations are possible but there is a finite 
number, depending on the number of steps. This is why there are so many lines in the atomic spectra — 
the electron can receive energy to promote it to a higher energy level and it can then fall to any level 
below and a certain quantity of light will be released. 

We want to predict, as far as we can, where all the electrons in different molecules are to be found 
including the ones not involved with bonding. We want to know where the molecule can accommo- 
date extra electrons and from where electrons can be removed most easily. Since most molecules 
contain many electrons, the task is not an easy one. However, the electronic structure of atoms is 
somewhat easier to understand and we can approximate the electronic structure of molecules by con- 
sidering how the component atoms combine. 

The next section is therefore an introduction to the electronic structure of atoms — what energies 
the electrons have and where they may be found. Organic chemists are rarely concerned with atoms 
themselves but need to understand the electronic structure in atoms before they can understand the 
electronic structure in molecules. As always, evidence first! 

The atomic emission spectrum of hydrogen 

The atomic emission spectrum of hydrogen is composed of many lines but these fall into separate 
sets or series. The first series to be discovered, not surprisingly, were those lines in the visible part of 
the spectrum. In 1885, a Swiss schoolmaster, Johann Balmer, noticed that the wavelengths, A,, of the 
lines in this series could be predicted using a mathematical formula. He did not see why; he just saw 
the relationship. This was the first vital step. 

n 2 
X = constant x (n is an integer greater than 2) 

n 2 -2 2 

As a result of his work, the lines in the visible spectrum are known as the Balmer series. The other 
series of lines in the atomic emission spectrum of hydrogen were discovered later (the next wasn't 
discovered until 1908). These series are named after the scientists who discovered them; for example, 
the series in the ultraviolet region is known as the Lyman series after Theodore Lyman. 

Balmer's equation was subsequently refined to give an equation that predicts the frequency, V, of 
any of the lines in any part of the hydrogen spectrum rather than just for his series. It turns out that 
his was not the most fundamental series, just the first to be discovered. 



constant x 



t \ 

1 1_ 

2 2 

n 1 n 2 



Each series can be described by this equation if a particular value is given for n\ but «2 is allowed to 
vary. For the Lyman series, n\ remains fixed at 1 while «2 can be 2, 3, 4, and so on. For the Balmer 
series, «i is fixed at 2 while H2 can be 3, 4, 5, and so on. 



Atomic structure 



85 



Atomic emission spectra are evidence for electronic energy levels 

Atomic emission spectra give us our first clue to understanding the electronic energy levels in an 
atom. Since the lines in the emission spectrum of hydrogen correspond to the electron moving 
between energy levels and since frequency is proportional to energy, E= hv, the early equations must 
represent just the difference between two energy levels. This in turn tells us that the electron's energy 
levels in an atom must be inversely proportional to the square of an important integer '«'. This can 
be expressed by the formula 

constant 

" = J~ 

where E n is the energy of an electron in the nth energy level and n is an integer > 1 known as the prin- 
cipal quantum number. Note that, when n = °°, that is, when the electron is no longer associated 
with the nucleus, its energy is zero. All other energy levels are lower than zero because of the minus 
sign in the equation. This is consistent with what we know already — we must put energy in to ionize 
the atom and remove the electron from the nucleus. 



Electronic energy levels 

In more detail, the constant in this equation can be broken 
down into a universal constant, the Rydberg constant R^, 
which applies to any electron on any atom, and a constant 
Zwhich has a particular value for each atom. 

RhZ 2 



The Rydberg constant Rh. is measured in units of energy. 
Foragiven atom (i.e. Zis constant) there are many 



different energy levels possible (each corresponding to a 
different value of n). Also, as ngets bigger, the energy gets 
smaller and smaller and approaches zero for large n. The 
energy gets smaller as the electron gets further away from 
the nucleus. For electrons in the same energy level but in 
different atoms, (i.e. keeping n constant but varying Z), the 
energy of an electron depends on the square of the atomic 
number. This makes sense too — the more protons in the 
nucleus, the more tightly the electron is held in the atom. 



• The electrons in any atom are grouped in energy levels whose energies are 

universally proportional to the inverse square of a very important number n. This 
number is called the principal quantum number and it can have only a few 
integral values (n = 1, 2, 3. . .)• The energy levels also depend on the type of atom. 

An energy level diagram gives some idea of the relative spacing between these energy levels. 
Principal quantum number, n 



E=0" 



n = 
n = 4 
r? = 3 



this diagram shows 

the spacing of the 

energy levels in a 

hydrogen atom 



this is the amount of 
energy needed to remove 
an electron in the lowest 
energy level completely 
away from the nucleus. 
It is the ionization 
energy of the atom 



4- 



a hydrogen atom in the 

ground state has one electron 

(represented by the arrow) 

in the lowest energy level 



Notice how the spacing between 
the energy levels gets closer and 
closer. This is a consequence of 
the energy being inversely 
proportional to the square of the 
principal quantum number. It tells 
us that it becomes easier and 
easier to remove an electron 
completely from an atom as the 
electron is located in higher and 
higher energy levels. As we shall 
see later, the increasing value of 
the principal quantum number 
also correlates with the electron 
being found (on average) further 
and further from the nucleus and 
being easier and easier to 
remove. This is analogous to a 
rocket escaping from a planet — 
the further away it is, the less it 
experiences the effects of gravity 
and so the less energy it requires 
to move still further away. The 
main difference is that there 
seems to be no quantization of 
the different energy levels of the 
rocket — it appears (to us in our 
macroscopic world at least) that 
any energy is possible. In the 
case of the electron in the atom, 
only certain values are allowed. 



86 



4 ■ Structure of molecules 



Three quantum numbers come from the Schrodinger equation 

There is no doubt about the importance of n, the principal quantum number, but where does it come 
from? This quantum number and two other quantum numbers come from solving the Schrodinger 
equation. We are not going to go into any details regarding Schrodinger's equation or how to solve 
it — there are plenty of more specialized texts available if you are interested in more detail. 

Solutions to Schrodinger's equation come in the form of wave functions (symbol V P), which 
describe the energy and position of the electrons thought of as waves. You might be a little unsettled 
to find out that we are describing electrons using waves but the same wave-particle duality idea 
applies to electrons as to light. We regularly think of light in terms of waves with their associated 
wavelengths and frequencies but light can also be described using the idea of photons — individual 
little light 'particles'. The same is true of the electron; up to now, you will probably have thought of 
electrons only as particles but now we will be thinking of them as waves. 

It turns out that there is not one specific solution to the Schrodinger equation but many. This is 
good news because the electron in a hydrogen atom can indeed have a number of different energies. 
It turns out that each wave function can be defined by three quantum numbers (there is also a fourth 
quantum number but this is not needed to define the wave function). We have already met the prin- 
cipal quantum number, n. The other two are called the orbital angular momentum quantum num- 
ber (sometimes called the azimuthal quantum number), £, and the magnetic quantum number, mi. 

A specific wave function solution is called an orbital. The different orbitals define different 
energies and distributions for the different electrons. The name 'orbital' goes back to earlier theories 
where the electron was thought to orbit the nucleus in the way that planets orbit the sun. It seems to 
apply more to an electron seen as a particle, and orbitals of electrons thought of as particles and wave 
functions of electrons thought of as waves are really two different ways of looking at the same thing. 
Each different orbital has its own individual quantum numbers, n, (,, and m%. 



Don't worry about the ratherfancy 
names of these quantum numbers; just 
accept that the three numbers define a 
given wave function. 



Summary of the importance of the quantum numbers 

What does each quantum number tell us and what values can it adopt? You have already met the 
principal quantum number, n, and seen that this is related to the energy of the orbital. 

The principal quantum number, n 

Different values for n divide orbitals into groups of similar energies called shells. Numerical values 
for n are used in ordinary speech. The first shell ( n = 1 ) can contain only two electrons and the atoms 
H and He have one and two electrons in this first shell, respectively. 

The orbital angular momentum quantum number, £ 

The orbital angular momentum quantum number, €, determines, as you might guess, the angular 
momentum of the electron as it moves in its orbital. This quantum number tells us the shape of the 
orbital, spherical or whatever. The values that € can take depend on the value of n: (, can have any 
value from up to n - 1: 1 = 0, 
1, 2, . . . . , n — 1, The different 
possible values of I are given 
letters rather than numbers 
and they are called s, p, d, and f. 

The magnetic quantum number, m^ 

The magnetic quantum number, mi, determines the 
spatial orientation of the angular momentum. In 
simple language it determines where the orbitals are 
in space. Its value depends on the value of t, varying 
from -i to +1: m t = i, i - 1, i - 2, . . . . , -i. The different 
possible values of m; are given suffixes on the letters 



value of n 


1 


2 


3 


4 


possible values of € 





0,1 


0,1,2 


0,1,2,3 


name 


Is 


2s, 2p 


3s, 3p,3d 


4s, 4p, 4d, 4f 



value of n 






1 


2 




2 


value of € 














1 


name 






Is 


2s 




2p 


possible val 


ues 


of me 








+1 


,0,-1 


name 






Is 


2s 


2Px. 


2p y , 2p z 



Atomic orbitals 



87 



defining the quantum number €. These letters refer to the direction of the orbitals along the x-, y-, or 
z-axes. Organic chemists are concerned mostly with s and p orbitals {(, = or 1) so the subdivisions 
of the d orbitals can be omitted. 

Each quantum number gives subdivisions for the one before. There are no subdivisions in the 
lowest value of each quantum number: and the subdivisions increase in number as each quantum 
number increases. Now we need to look in more detail at the meanings of the various values of the 
quantum numbers. 



Atomic orbitals 

Nomenclature of the orbitals 

For a hydrogen atom the energy of the orbital is determined only by the principal quantum number, 
n, and n can take values 1, 2, 3, and so on. This is the most fundamental division and is stated first in 
the description of an electron. The electron in a hydrogen atom is called Is . The 1 gives the value of 
n: the most important thing in the foremost place. The designation s refers to the value of €, These 
two together, Is, define and name the orbital. The superscript 1 tells us that there is one electron in 
this orbital. 

The orbital angular momentum quantum number, €, determines the shape of the orbital. Instead 
of expressing this as a number, letters are used to label the different shapes of orbitals. s orbitals have 
€ = 0, and p orbitals have € = 1. 

Using both these quantum numbers we can label orbitals Is, 2s, 2p, 3s, 3p, 3d, and so on. Notice 
that, since I can only have integer values up to n - 1 , we cannot have a lp or 2d orbital. 

• Names of atomic orbitals 

• The first shell (n = 1) has only an s orbital, Is 

• The second shell (n = 2) has s and p orbitals 2s and 2p 

• The third shell (n = 3) has s, p, and d orbitals, 3s, 3p, and 3d 

One other point to notice is that, for the hydrogen atom (and, technically speaking, any one-elec- 
tron ion such as He + or Li + ), a 2s orbital has exactly the same energy as a 2p orbital and a 3s orbital 
has the same energy as the 3p and 3d orbitals. Orbitals that have the same energy are described as 
degenerate. In atoms with more than one electron, things get more complicated because of elec- 
tron-electron repulsion and the energy levels are no longer determined by n alone. In such cases, the 
2s and 2p or the 3s, 3p, and 3d orbitals or any other orbitals that share the same principal quantum 
number are no longer degenerate. In other words, in multielectron atoms, the energy of a given 
orbital depends not only on the principal quantum number, n, but also in some way on the orbital 
angular momentum quantum number, (,. 

Values of the magnetic quantum number, m;, depend on the value of (,. When € = 0, mg can only 
take one value (0); when £ = 1, m/has three possible values (+1, 0, or -1). There are five possible val- 
ues of m; when € = 2 and seven when £ = 3. In more familiar terms, there is only one sort of s orbital; 
there are three sorts of p orbitals, five sorts of d orbitals, and seven sorts of f orbitals. All three p 
orbitals are degenerate as are all five d orbitals and all seven f orbitals (for both single-electron and 
multielectron atoms). We shall see how to represent these orbitals later. 

There is a fourth quantum number 

The spin of an electron is the angular momentum of an electron spinning about its own 
axis, although this is a simplified picture. This angular momentum is different from the 
angular momentum, €, which represents the electron's angular momentum about the nucleus. The 
magnitude of the electron's spin is constant but it can take two orientations. These are represented 
using the fourth quantum number, the spin angular momentum quantum number, m s , which 
can take the value of +1 or -1 in any orbital, regardless of the values of n, £, or m^. Each 



s, p, d, f 

These letters hark back to the early 
days of spectroscopy and refer to 
the appearance of certain lines in 
atomic emission spectra: 's' for 
'sharp', 'p' for 'principal', 'd' for 
'diffuse', and T for 'fundamental'. 
The letters s, p, d, and f matter and 
you must know them, but you do 
not need to know what they 
originally stood for. 



Value of / 


Name of orbital 





s 


1 


P 


2 


d 


3 


f 



You have already come across another 
spin — the nuclear spin — which gives 
rise to NMR. There is an analogous 
technique, electron spin resonance, 

ESR, which detects unpaired electrons. 



88 



4 ■ Structure of molecules 



orbital can hold a maximum of two electrons and then only when the electrons have different 
'spin', that is, they must have different values of m s , +1 or -1. The rule that no more than two 
electrons may occupy any orbital (and then only if their spins are paired) is known as the Pauli 
exclusion principle. 



Every electron is unique! 

If electrons are in the same atom, they must have a unique combination of the four 
quantum numbers. Each orbital, designated by three quantum numbers, n, €, and 
m%, can contain only two electrons and then only if their spin angular quantum 
numbers are different. 



How the periodic table is constructed 

All the quantum numbers for all the electrons with « = 1 and 2 can now be shown in a table like the 
ones earlier in this chapter. Though we have so far been discussing the hydrogen atom, in fact, the H 
atom never has more than two electrons. Fortunately, the energy levels deduced for H also apply to 
all the other elements with some minor adjustments. This table would actually give the electronic 
configuration of neon, Ne. 

In this table, the energy goes up 
from left to right, though all the 2p 
orbitals are degenerate. To add n = 
3, one column for the 3s, three 
columns for the 3p, and five 
columns for the 3d orbital would 
be needed. Then all five 3d orbitals 
would be degenerate. 

Another way to show the same thing is by an energy diagram showing how the quantum numbers 
divide and subdivide. 



value of n 


1 


2 


2 


2 


2 


value of t 








1 


1 


1 


name 


Is 


2s 


2p 


2p 


2p 


possible values of mi 








+1 





-1 


name 


Is 


2s 


2Px 


2p y 


2P Z 


possible values of m s 


+1,-1 


+1,-1 


+1,-1 


+1,-1 


+1,-1 


electrons 


Is 2 


2s 2 


2p 2 


2PJ 


2P 2 



/ 



2p 





2s 



m, 



2Pz 



- o:: 

2p y 



-1: 

2p„ 



0: 

2s 



m s 
+1 



+1 



+1 



+1 



six 2p 
electrons 



two 2s 
electrons 



8 electrons 
in n = 2 shell 



-0 

Is 



0: 

Is 



+1 



two Is 
electrons 



2 electrons 
in n = 1 shell 



Atomic orbitals 



89 



These numbers explain the shape of the periodic table. Each element has one more electron (and 
one more proton and perhaps more neutrons) than the one before. At first the lowest energy shell (« 
= 1) is filled. There is only one orbital, Is, and we can put one or two electrons in it. There are there- 
fore two elements in this block, H and He. Next we must move to the second shell (« = 2), filling 2s 
first so we start the top of groups 1 and 2 with Li and Be. These occupy the top of the red stack 
marked 's block' because all the elements in this block have one or two electrons in their outermost s 
orbital and no electrons in the outermost p orbital. Then we can start on the 2p orbitals. There are 
three of these so we can put in six electrons and get six elements B, C, N, O, F, and Ne. They occupy 
the top row of the black p block. Most of the elements we need in this book are in those blocks. Some, 
Na, K, and Mg for example, are in the s block and others, Si, P, and S for example, are in the second 
row of the p block. 

The layout of the periodic table 

s block - 
each s orbital 
can hold only 
two electrons ± s 



2s 






3s 






4s 






5s 






6s 






7s 







p block - each set of p orbitals 
can hold six electrons 



d block - each set of d orbitals can hold ten electrons 3p 



3d 






















4d 






















5d 






















6d 
















! ! ' 



2p 














3p 














4p 














5p 














6p 



















f block ■ 


- eac 


l set of f orbitals can 


iold fourteen electrons 




4f 






























5f 































Other orbitals 

Organic chemists are really 
concerned only with s and p 
orbitals since most of the 
elements we deal with are in the 
second row of the periodic table. 
Later in the book we shall meet 
elements in the second row of the 
p block (Si, P, S) and then we will 
have to consider their d orbitals, 
but for now we are not going to 
bother with these and certainly not 
with the f orbitals. But you may 
have noticed that the 4s orbital is 
filled before the 3d orbitals so you 
may guess that the 4s orbital must 
be slightly lower in energy than the 
3d orbitals. Systems with many 
electrons are more complicated 
because of electron repulsion and 
hence the energies of theirorbitals 
do not simply depend on n alone. 



Graphical representations of orbitals 

One problem with wave functions is trying to visualize them: what does a wave function look like? 
Various graphs of wave functions can be plotted but they are not much help as "P itself has no physi- 
cal meaning. However the square of the wave function, "P , does have a practical interpretation; it is 
proportional to the probability of finding an electron at a given point. Unfortunately, we can't do 



' \ A, / \ / 

■** — V ■ + *- \ / 

-« »~ 



These two waves both have the same wavelength, X, but 
the dashed wave is less intense than the other wave. 
The intensity is proportional to the amplitude squared. 



There is some justification forthis 
interpretation that the wave 
function squared is proportional 
to the probability of finding an 
electron. With lightwaves, for 
example, while the wavelength 
provides the colour (more 
precisely the energy) of the wave, 
it is the amplitude squared that 
gives the brightness. 

But this is looking at light in terms 
of waves. In terms of particles, 
photons, the intensity of light is 
proportional to the density of 
photons. 



90 



4 ■ Structure of molecules 



better than probability as we are unable to say exactly where the electron is at any time. This is a con- 
sequence of Heisenberg's uncertainty principle — we cannot know both the exact position and the 
exact momentum of an electron simultaneously. Here we know the momentum (energy) of the elec- 
tron and so its exact position is uncertain. 

How do we depict a probability function? One way would be to draw contours connecting regions 
where there is an equal probability of finding the electron. If *P for a Is orbital is plotted, a three- 
dimensional plot emerges. Of course, this is a two-dimensional representation of a three-dimension- 
al plot — the contours are really spherical like the different layers of an onion. These circles are rather 
like the contour lines on a map except that they represent areas of equal probability of finding the 
electron instead of areas of equal altitude. 

Another way to represent the probability is by a density plot. Suppose we could see exactly where 
the electron was at a given time and that we marked the spot. If we looked again a little later, the elec- 
tron would be in a different place — let us mark this spot too. Eventually, if we marked enough spots, 
we would end up with a fuzzy picture like those shown for the Is and 2s orbitals. Now the density of 
the dots is an indication of the probability of finding an electron in a given space — the more densely 
packed the dots (that is, the darker the area), the greater the probability of finding the electron in this 
area. This is rather like some maps where different altitudes are indicated by different colours. 




contour diagram of Is orbital 





density plot of Is orbital 



density plot of 2s orbital 



The 2s orbital, like the Is orbital, is spherical. There are two differences between these orbitals. 
One is that the 2s orbital is bigger so that an electron in a 2s orbital is more likely to be found further 
away from the nucleus than an electron in a Is orbital. The other difference between the orbitals is 
that, within the 2s orbital but not within the Is orbital, there is a region where there is no electron 
density at all. Such a region is called a nodal surface. In this case there is no electron density at one set 
radius from the nucleus; hence this is known as a radial node. The 2s orbital has one radial node. 



Nodes are important for musicians 

You can understand these nodal surfaces by thinking in terms of waves. If a 
violin or other string instrument is plucked, the string vibrates. The ends 
cannot move since they are fixed to the instrument. The note we hear is 
mainly due to the string vibrating as shown in the diagram for the first 

harmonic. 

no no< 
However, there are other vibrations of higher energy 
known as harmonics, which help to give the note its 
timbre (the different timbres allow us to tell the 
difference between, say, a flute and a violin playing 
the same note). The second and third harmonics are 
also shown. 

Each successive harmonic has one extra node — while 
the first harmonic has no nodes (if you don't count the 



end stops), the second harmonic has one and the third has two and so on. 
These are points where the string does not vibrate at all (you can even 'select 
out' the second harmonic on a stringed instrument if you gently press halfway 
along the vibrating string). 



1 node 





2 nodes 



J/.-M 




1st harmonic 



2nd harmonic 



3rd harmonic 



Atomic orbitals 



91 



• Shapes of s orbitals 




• The Is orbital is spherically symmetrical and has no nodes 


• The 2s orbital has one radial node and the 


3s orbital two radial nodes. 


They are both spherically symmetrical 





What does a f for a 2p orbital look like? The probability density plot is no longer spherically 
symmetrical. This time the shape is completely different — the orbital now has an orientation in 
space and it has two lobes. Notice also that there is a region where there is no electron density 
between the two lobes — another nodal surface. This time the node is a plane in between the two 
lobes and so it is known as a nodal plane. One representation of the 2p orbitals is a three-dimension- 
al plot, which gives a clear idea of the true shape of the orbital. 



• 




density plot of 2p orbital 



three-dimensional plot of the 2p orbitals 



Plots of 3p and 4p orbitals are similar — each has a nodal plane and the overall shape outlined in 
each is the same. However, the 3p orbital also has a radial node and the 4p has two radial nodes and 
once again the size of the orbital increases as the principal quantum number increases. 

All this explains why the shape of an orbital depends on the orbital angular quantum number, (,. 
All s orbitals (€= 0) are spherical, all p orbitals (€ = 1) are shaped like a figure eight, and d orbitals 
(€= 2) are yet another different shape. The problem is that these probability density plots take a long 
time to draw — organic chemists need a simple easy way to represent orbitals. The contour diagrams 
were easier to draw but even they were a little tedious. Even simpler still is to draw just one contour 
within which there is, say, a 90% chance of finding the electron. This means that all s orbitals can be 
represented by a circle, and all p orbitals by a pair of lobes. 

The phase of an orbital 

The wave diagrams need further discussion to establish one fine point — the phase of an orbital. 

after the node, the phase 1 node 
of the wave changes 

above this line the phase 
**. of the wave is positive 




You might have noticed that each 
orbital in the nth energy level has 
the same total number of nodes, 
n-1. The total number of nodes 
is the sum of the numbers of 
radial nodes and nodal planes. 
Thus both the 2s and 2p have one 
node (a radial node in the case of 
the 2s and a nodal plane in the 
case ofthe2p) while the 3s, 3p, 
and 3d orbitals each have two 
nodes (the 3s have two radial 
nodes, the 3p orbitals each have 
one radial node and one nodal 
plane, and the 3d orbitals each 
have two nodal planes). 




an s orbital 




a p orbital 



Remember that the orbitals are three- 
dimensional and that these drawings 
represent a cross-section. Athree- 
dimensional version would look more 
like a sphere for an s orbital and an old- 
fashioned hour-glass for a p orbital. 
Actually, each lobe of a p orbital is 
much more rounded than the usual 
representation, but that is not so 
important. 



below the line, the phase 
of the wave is negative 



Just as an electromagnetic wave, or the wave on a vibrating string, or even an ocean wave possess- 
es different 'phases' (for example, the troughs and peaks of an ocean wave) so too do the atom's wave 
functions — the orbitals. After each node in an orbital, the phase of the wave function changes. In the 



92 



4 ■ Structure of molecules 



2p orbital, for example, one lobe is one phase; the other lobe is another phase with the nodal plane in 
between. In the standing wave above the different phases are labelled positive and negative. The 
phases of a p orbital could be labelled in the same way (and you may sometimes see this) but, since 
chemists use positive and negative signs to mean specific charges, this could get confusing. Instead, 
one half of the p orbital is usually shaded to show that it has a different phase from the other half. 



here the different 

phases of the p 

orbital are labelled 

positive and negative - 

this can be confusing 

and so is best avoided 



here the different 
phases of the p 
orbital are shown 
by shading one half 
and not the other 



The magnetic quantum number, nt£ 

The magnetic quantum number, m/, determines the spatial orientation of the orbital's angular 
momentum and takes the values —i to +€, An s orbital (€ = 0), being spherical, can only have one 
orientation in space — it does not point in any one direction and hence it only has one value for 
ni((0). However, a p orbital could point in any direction. For a p orbital (€ = 1 ) there are three values 
of m/: -1, 0, and +1. These correspond to the p orbitals aligned along the mutually perpendicular x-, 
y-, and z-axes. These orbitals, designated p„ p-p and p z , are all degenerate. They differ only in their 
spatial orientations. 




x x 

the three degenerate p orbitals are aligned along perpendicular axes 



Summary so far 

• Electrons in atoms are best described as waves 

• All the information about the wave (and hence about the electron) is in the wave function, 'P, the 
solution to the Schrodinger equation 

• There are many possible solutions to the Schrodinger equation but each wave function (also 
called an orbital) can be described using three quantum numbers 

• The principal quantum number, n, is largely responsible for the energy of the orbital (in one- 
electron systems, such as the hydrogen atom, it alone determines the energy). It takes integer 
values 1, 2, 3, 4, and so on, corresponding to the first, second, third, and so on shells of electrons 

• The orbital angular momentum quantum number, €, determines the angular momentum that 
arises from the motion of an electron moving in the orbital. Its value depends on the value of n 
and it takes integer values 0, . . ., n— 1 but the orbitals are usually known by letters (s when € = 0, p 
when € = 1, d when € = 2, and f when € = 3). Orbitals with different values of € have different 
shapes — s orbitals are spherical, p orbitals are shaped like a figure of eight 

• The magnetic quantum number, trig, determines the spatial orientation of the orbital. Its value 

depends on the value of € and it can take the integer values: -€, 0, +(,. This means that 

there is only one type of s orbital, three different p orbitals (all mutually perpendicular), five dif- 
ferent d orbitals, and seven different f orbitals. The three different p orbitals are all degenerate, 
that is, they have the same energy (as do the five d orbitals and the seven f orbitals) 



Atomic orbitals 93 



• There is also a fourth quantum number, the spin angular momentum quantum number, Wj, 
which can take values of + 1 or -1. The spin is not a property of orbitals but of the electrons that 
we put in the orbitals 

• No two electrons in any one atom can have all four quantum numbers the same — this means that 
each orbital as described by the (first) three quantum numbers can hold a maximum of two elec- 
trons and then only if they have opposing spins 

• We usually use a shorthand notation to describe an orbital such as Is or 2p„ 

the number tells us the principal quantum number, n 

2s 3p x 4p y 

the letter tells us the the subscript letter 

orbital angular momentum tells us the magnetic 
quantum number, / quantum numer, m, 

these three quantum numbers, n, I, and m h define an orbital 

A few points are worth emphasis. Orbitals do not need to have electrons in them — they can be 
vacant (there doesn't have to be someone standing on a stair for it to exist!). So far we have mainly 
been talking about the hydrogen atom and this has only one electron. Most of the time this electron 
is in the Is orbital (the orbital lowest in energy) but if we give it enough energy we can promote it to 
a vacant orbital higher in energy, say, for example, the 3p x orbital. 

Another point is that the electrons may be found anywhere in an orbital except in a node. In a p 
orbital containing one electron, this electron may be found on either side but never in the middle. 
When the orbital contains two electrons, one electron doesn't stay in one half and the other electron 
in the other half — both electrons could be anywhere (except in the node). 

Finally, remember that all these orbitals are superimposed on each other. The Is orbital is not the 
middle part of the 2s orbital. The Is and 2s orbitals are separate orbitals in their own rights and each 
can hold a maximum of two electrons but the 2s orbital does occupy some of the same space as the Is 
orbital (and also as the 2p orbitals, come to that). Neon, for example, has ten electrons in total: two 
will be in the Is orbital, two in the 2s orbital, and two in each of the 2p orbitals. All these orbitals are 
superimposed on each other but the pairs of electrons are restricted to their individual orbitals. If we 
tried to draw all these orbitals, superimposed on each other as they are, in the same diagram the 
result would be a mess! 

Putting electrons in orbitals 

Working out where the electrons are in any atom, that is, which orbitals are populated, is easy. We 
simply put two electrons into the lowest energy orbital and work upwards. This 'building up' of the 
different atoms by putting electrons in the orbitals until they are full and then filling up the orbital 
next lowest in energy is known as the Aufbau principle (Aufbau is German for 'building up'). The 
first and only electron in the hydrogen atom must go into the Is orbital. In this sort of diagram the 
energy levels are represented as horizontal lines stacked roughly in order with the lowest energy at 
the bottom. Electrons are represented as vertical arrows. Arrows pointing upwards show one spin 
(m s = +1 or -1) and arrows pointing downwards the other (which is which doesn't matter). 

the 2s and 2p orbitals for hydrogen are 
degenerate. This is because this is a one- 
2s 2d 2n 2d electron system and the energy of the orbitals 

depends only on the principal quantum number, n 



this arrow represents an electron in the lowest (Is) 
-orbital. The direction of the arrow indicates the 
electron's spin and may be either up or down 



Is 

energy level diagram for a hydrogen atom (atomic number = 1) 



94 



4 ■ Structure of molecules 



The helium atom has two electrons and they can both fit into the Is orbital providing they have oppo- 
site spins. The other change to the diagram is that, with two electrons and electron repulsion a factor, the 
2s orbital is now lower in energy than the three 2p orbitals, though these three are still degenerate. 



This is known as Hund's rule. An 

atom adopts the electronic 
configuration that has the 
greatest number of unpaired 
electrons in degenerate orbitals. 
Whilst this is all a bit theoretical in 
that isolated atoms are not found 
very often, the same rule applies 
for electrons in degenerate 
orbitals in molecules. 



2s 



2Px 2p y 2p z 



Is 



the 2s and 2p orbitals for hydrogen are 
degenerate. This is because this is a one- 
electron system and the energy of the orbitals 
depends only on the principal quantum number, n 



this arrow represents an electron in the lowest (Is) 
-orbital. The direction of the arrow indicates the 
electron's spin and may be either up or down 



energy level diagram for a hydrogen atom (atomic number = 1) 
Lithium has one more electron but the Is orbital is already full. The third electron must go into 
the next lowest orbital and that is the 2s. In this three-electron system, like that of the two-electron 
He atom, the three 2p orbitals are higher in energy than the 2s orbital. By the time we come to boron, 
with five electrons, the 2s is full as well and we must put the last electron into a 2p orbital. It doesn't 
matter which one; they are degenerate. 

this is the same energy 
level diagram as for helium 
with more electrons in it 



2Px 



2p y 

2s 

Is 



2p z 



the two electrons in 
each of the Is and 2s 
orbitals have 
opposite spins - 
they are spin paired 



energy level diagram for a boron atom (atomic number = 5) 

Carbon has one more electron than boron but now there is a bit of a problem — where does the 
last electron go? It could either be paired with the electron already in one of the p orbitals or it could 
go into one of the other degenerate p orbitals. It turns out that the system is lower in energy (elec- 
tron-electron repulsion is minimized) if the electrons are placed in different degenerate orbitals with 
their spins parallel (that is, both spins +1 or both -1). Another way of looking at this is that putting 
two electrons into the same orbital with their spins paired (that is, one +1, one -1) requires some 
extra amount of energy, sometimes called pairing energy. 



here the electrons are paired 

4h - -••- 

2p x \ 2p y / 2p z 



here the electrons are in different degenerate 
2p orbitals with their spins parallel 

4--t 



.'2 



2p x 2p y 2p 2 

-H- 



-H- \ 



Is 
not observed since higher in energy 



2s 



Is 

lower in energy and the one the 
carbon atom actually adopts 



the two possible arrangements for the electrons in a carbon atom 



Molecular orbitals — homonuclear diatomics 



95 



Nitrogen, with one more electron than carbon, has a single electron in each of the 2p orbitals. The 
new electron pairs up with another already in one of the 2p orbitals. It doesn't enter the 3s orbital 
(the orbital next lowest in energy) since this is so much higher in energy and to enter the 3s orbital 
would require more energy than that needed to pair up with a 2p electron. 



3s 



"T t f" 



2p x 2p y 2p z 



2s 



Is 

nitrogen atoms have three 
unpaired electrons 



3s 



2s 



3s 



2p x 2p y 2p z 

-u- 



-f- J \fr -f- 

2p x 2i>y 2p z 



Is 

oxygen atoms have 
two unpaired electrons 



2s 



Is 

this arrangement of the electrons in 
an oxygen atom is higher in energy 



Molecular orbitals — homonuclear diatomics 

So far the discussion has concerned only the shapes and energies of atomic orbitals (AOs). Organic 
chemists really need to look at the orbitals for whole molecules. One way to construct such molecu- 
lar orbitals (MOs) is to combine the atomic orbitals of the atoms that make up the molecule. This 
approach is known as the Linear Combination of Atomic Orbitals (LCAO). 

Atomic orbitals are wave functions and the different wave functions can be combined together 
rather in the way waves combine. You may be already familiar with the ideas of combining waves — 
they can add together constructively (in-phase) or destructively (out-of-phase). 




combine 
in-phase 




constructive overlap 




combine 
out-of-phase 



destructive overlap 



the two ways of combining a simple wave - in-phase and out-of-phase 

Atomic orbitals can combine in the same way — in-phase or out-of-phase. Using two Is orbitals 
drawn as circles (representing spheres) with dots to mark the nuclei and shading to represent phase, 
we can combine them in-phase, that is, add them together, or out-of-phase when they cancel each 
other out in a nodal plane down the centre between the two nuclei. The resulting orbitals belong to 
both atoms — they are molecular rather than atomic orbitals. As usual, the higher energy orbital is at 
the top. 



96 



4 ■ Structure of molecules 



nodal plane 




Q 



combine 



o 



out-of-phase 
the two Is orbitals combining out-of-phase to give an antibonding orbital 

combine 






in-phase 



the two Is orbitals combining in-phase to give a bonding orbital 



When the two orbitals combine out-of-phase, the resulting molecular orbital has a nodal plane 
between the two nuclei. This means that if we were to put electrons into this orbital there would be 
no electron density in between the two nuclei. By contrast, if the molecular orbital from in-phase 
combination contained electrons, they would be found in between the two nuclei. Two exposed 
nuclei repel each other as both are positively charged. Any electron density between them helps to 
bond them together. So the in-phase combination is a bonding molecular orbital. As for the elec- 
trons themselves, they can now be shared between two nuclei and this lowers their energy relative to 
the Is atomic orbital. Electrons in the orbital from the out-of-phase combination do not help bond 
the two nuclei together; in fact, they hinder the bonding. When this orbital is occupied, the electrons 
are mainly to be found anywhere but between the two nuclei. This means the two nuclei are more 
exposed to each other and so repel each other. This orbital is known as an antibonding molecular 
orbital and is higher in energy than the Is orbitals. 

The combination of the atomic Is orbitals to give the two new molecular orbitals is simply shown 
on an energy level diagram. With one electron in each Is orbital, two hydrogen atoms combine to 
give a hydrogen molecule. 

(empty) antibonding molecular orbital 



combine 
out-of-phase 



o o 



O 



-O 



Is 
atomic 
orbital 


combine 




in-phase 


hydrogen 
atom A 





<TD 



Is 

atomic 
orbital 



hydrogen 
atom B 



(full) bonding molecular orbital 
the hydrogen molecule resulting from the combination of the two hydrogen atoms 



There are several points to notice about this diagram. 

Two atomic orbitals (AOs) combine to give two molecular orbitals (MOs) 

By LCAO we add the two AOs to make the bonding orbital and subtract them to make the anti- 
bonding orbital 

Since the two atoms are the same, each AO contributes the same amount to the MOs 

The bonding MO is lower in energy than the AOs 

The antibonding MO is higher in energy than the AOs 

Each hydrogen atom initially had one electron. The spin of these electrons is unimportant 



Molecular orbitals — homonuclear diatomics 



97 



• The two electrons end up in the MO lowest in energy. This is the bonding MO 

• Just as with AOs, each MO can hold two electrons as long as the electrons are spin paired 

• The two electrons between the two nuclei in the bonding MO hold the molecule together — they 
are the chemical bond 

• Since these two electrons are lower in energy in the MO than in the AOs, energy is given out when 
the atoms combine 

• Or, if you prefer, we must put in energy to separate the two atoms again and to break the 
bond 

From now on, we will always represent molecular orbitals in energy order — the highest-energy 
MO at the top (usually an antibonding MO) and the lowest in energy (usually a bonding MO and the 
one in which the electrons are most stable) at the bottom. We suggest you do the same. 

When we were looking at the electronic configuration of atoms, we simply filled up the atomic 
orbitals starting from the lowest in energy and worked up. With molecules we do the same: we just 
fill up the molecular orbitals with however many electrons we have, starting from the lowest in 
energy and remembering that each orbital can hold two electrons and then only if they are spin 
paired. 

Breaking bonds 

If an atom is supplied with energy, an electron can be promoted to a higher energy level and it can 
then fall back down to its ground state, giving that energy out again. What would happen if an elec- 
tron were promoted in a hydrogen molecule from the lowest energy level, the bonding MO, to the 
next lowest energy level, the antibonding MO? Again, an energy level diagram helps. 



. — - ^__ antibonding 

' -* ^«J molecular orbital 



AO 



energy 

needed to 

promote 

electron 



AO 



cn> 



bonding 
molecular 
orbital 



we can supply energy to promote an electron from the bonding MO to the antibonding MO 



Now the electron in the antibonding orbital 'cancels out' the bonding of the electron in the bond- 
ing orbital. Since there is no overall bonding holding the two atoms together, they can drift apart as 
two separate atoms with their electrons in Is atomic orbitals. In other words, promoting an electron 
from the bonding MO to the antibonding MO breaks the chemical bond. This is difficult to do with 
hydrogen molecules but easy with, say, bromine molecules. Shining light on Br2 causes it to break up 
into bromine atoms. 

Bonding in other elements: helium 

A hydrogen molecule is held together by a single chemical bond since the pair of electrons in the 
bonding orbital constitutes this single bond. What would the MO energy level diagram for He2 look 
like? Each helium atom has two electrons (Is ) so now both the bonding MO and the antibonding 
MO are full. Any bonding due to the electrons in the bonding orbital is cancelled out by the electrons 
in the antibonding orbital. 



This idea will be developed in Chapters 
5 and 6 when we look at bond-breaking 
steps in organic reaction mechanisms. 



98 



4 ■ Structure of molecules 



O 



(full) antibonding molecular orbital 



combine 
out-of-phase 



o o 



Is 


-.. 


atomic 


*^ 


orbital 


combine 




in-phase 


helium 




atom A 





d^> 



-o 

Is 
atomic 
orbital 



helium 
atom B 



(full) bonding molecular orbital 
the hypothetical molecule resulting from the combination of the two helium atoms 



There is no overall bonding, the two helium atoms are not held together, and He2 does not exist. 
Only if there are more electrons in bonding MOs than in antibonding MOs will there be any bonding 
between two atoms. In fact, we define the number of bonds between two atoms as the bond order 
(dividing by two since two electrons make up a chemical bond). 



bond order 



(no. of electrons in bonding MOs)- (no. of electrons in antibonding MOs) 



Hence the bond orders for H 2 and He2 are 
2-0 



bond order (H 2 ) 



1 i.e. a single bond 



bond order (He 2 ) = 



2-2 



i.e. no bond 



Antibonding orbitals are designated 
with a * e.g. o*,orji* 



Bond formation using 2s and 2p atomic orbitals 

So far we have been looking at how we can combine the Is atomic orbitals to give the molecular 
orbitals of simple molecules. However, just as there are lots of higher, vacant energy levels in atoms, 
so there are in molecules too. Other atomic orbitals combine to give new molecular orbitals and the 
2s and 2p orbitals concern organic chemistry most of all. The 2s AOs combine in exactly the same 
way as the Is orbitals do and also give rise to a bonding and an antibonding orbital. With p orbitals as 
well, there are more possibilities. 

Since we are beginning to talk about lots of different MOs, we shall need to label them with a little 
more thought. When s orbitals combine, the resulting MOs, both bonding and antibonding, are 
totally symmetrical about the axis joining the two nuclei. 



-e>~o-e 



we can rotate 
about this axis 
without changing 
the MOs 



both MOs have rotational symmetry about the axis through the two nuc 




When orbitals combine in this end-on overlap to give cylindrically symmetrical MOs, the result- 
ing orbitals are said to possess sigma (o) symmetry. Hence the bonding MO is a sigma orbital and 
electrons in such an orbital give rise to a sigma bond. In the hydrogen molecule the two hydrogen 
atoms are joined by a bond. 



Molecular orbitals — homonuclear diatomics 



99 



What MOs result from the combina- these two pairs of p orbitals must combine side-on 

tion of two p orbitals? There are three 
mutually perpendicular p orbitals on each 
atom. As the two atoms approach each 
other, these orbitals can combine in two 
different ways — one p orbital from each 
atom can overlap end-on, but the other 
two p orbitals on each atom must com- 
bine side-on. 

The end-on overlap (in-phase and out- 
of-phase) results in a pair of MOs that are 
cylindrically symmetrical about the inter- 
nuclear axis — in other words, these combinations have o" symmetry. The two molecular orbitals 
resulting from the end-on combination of two 2p orbitals are labelled the 2p0 and the 2po"* MOs. 





only these two p orbitals can overlap end-on 



two different ways that p orbitals can overlap with each other 



OOCXD 



combine 



noda| plane 



^> 



2p AO 



2p AO 



out-of-phase 



O-jO-e 

2po* MO 



symmetrical 
about this axis. 



the end-on overlap of two 2p atomic orbitals to give the 2po~ antibonding MO 



oooo 



2pA0 



2p AO 



combine 



in-phase 



2pa MO 



the end-on overlap of two 2p atomic orbitals to give the 2po bonding MO 



-e- 



symmetrical 
about this axis. 



The side-on overlap of two p orbitals forms an MO that is no longer symmetrical about the inter- 
nuclear axis. If we rotate about this axis, the phase of the orbital changes. The orbital is described as 
having Jt symmetry — a 7C orbital is formed and the electrons in such an orbital make up a 7C bond. 
Since there are two mutually perpendicular pairs of p orbitals that can combine in this fashion, there 
are a pair of degenerate mutually perpendicular % bonding MOs and a pair of degenerate mutually 
perpendiculars* antibonding MOs. 



combine 



out-of-phase 



nodal plane 



2p AO 2p AO 2p7C* MO 

the side-on overlap of two 2p atomic orbitals to give the 2pji* antibonding MO 



e 



no symmetry 

about this axis. 

If we rotate, 

the phase changes 



combine 



in-phase 



CO 



2p AO 2p AO 2pji MO 

the side-on overlap of two 2p atomic orbitals to give the 2pn bonding MO 



e 



no symmetry 

about this axis. 

If we rotate, 

the phase changes 



The two sorts of molecular orbitals arising from the combinations of the p orbitals are not degen- 
erate — more overlap is possible when the AOs overlap end-on than when they overlap side-on. As a 



100 



4 ■ Structure of molecules 



result, the p0 orbital is lower in energy than the pJt orbital. We can now draw an energy level diagram 
to show the combination of the Is, 2s, and 2p atomic orbitals to form molecular orbitals. 



Homonuclearand heteronuclear 
referto the nature of the atoms in 
a diatomic molecule. In a 
homonuclear molecule the atoms 
are the same (such as H 2 , N 2 , 2 , 
F 2 ) while in a heteronuclear 
molecule they are different (as in 
HF, CO, NO, ICI). 



2po* 



3x 2p 



2 x 2pjf 
< 



2 x 2pn 



3 x 2p 



2 pa 
2sa* 



2s 



2s 



2 so 

the lso and lso* MOs are much lower in energy than the other MOs 
lsa* 



Is 



atomic orbitals 
on atom A 



lso 



Is 



atomic orbitals 
on atom B 



molecular orbitals resulting from the combination of atomic orbitals 

Let us now look at a simple diatomic molecule — nitrogen. A nitrogen molecule is composed of 
two nitrogen atoms, each containing seven electrons in total. We shall omit the Is electrons because 
they are so much lower in energy than the electrons in the 2s and 2p AOs and because it makes no 
difference in terms of bonding since the electrons in the ls0* cancel out the bonding due to the elec- 
trons in the ls0 MO. The electrons in the Is AOs and the Is MOs are described as core electrons and 
so, in discussing bonding, we shall consider only the electrons in the outermost shell, in this case the 
2s and 2p electrons. This means each nitrogen contributes five bonding electrons and hence the mol- 
ecular orbitals must contain a total often electrons. 

The electrons in the and 0* MOs formed 
from the 2s MOs also cancel out — these electrons 
effectively sit on the atoms, two on each, and form 
lone pairs — nonbonding pairs of electrons that do 
not contribute to bonding. All the bonding is done 
with the remaining six electrons. They fit neatly 
into a O bond from two of the p orbitals and two 7t 
bonds from the other two pairs. Nitrogen has a 
triple bonded structure. 



nonbonding 
lone pair 



two ji bonds 



t 
N=N : -*- 



one o bond 



nonbonding 
lone pair 



Heteronuclear diatomics 

Up to now we have only considered combining two atoms of the same element to form homonuclear 
diatomic molecules. Now we shall consider what happens when the two atoms are different. First of 
all, how do the atomic orbitals of different elements differ? They have the same sorts of orbitals Is, 2s, 
2p, etc. and these orbitals will be the same shapes but the orbitals will have different energies. For 



Heteronuclear diatomics 



101 



example, removing an electron completely from atoms of carbon, oxygen, or fluorine (that is, 
ionizing the atoms) requires different amounts of energy. Fluorine requires most energy, carbon 
least, even though in each case we are removing an electron from the same orbital, the 2p AO. The 
energies of the 2p orbitals must be lowest in fluorine, low in oxygen, and highest in carbon. 

£=0 - 



less 

energy needed 

to ionize a 

carbon atom 



4- 



4- 



2p x 2p y 2p z 



2s 



atomic orbitals for carbon 



energy needed 
to ionize an 
oxygen atom 



more 

energy needed 

to ionize a 

fluorine atom 



-H-4-4 



2Px 2p y 2p z 



2s 



atomic orbitals for oxygen 



A l kf> 

2p* 2p y 2p 



2s 



atomic orbitals for fluorine 



We are talking now about electronegativity. The more electronegative an atom is, the more it 
attracts electrons. This can be understood in terms of energies of the AOs. The more electronegative 
an atom is, the lower in energy are its AOs and so any electrons in them are held more tightly. This is 
a consequence of the increasing nuclear charge going from left to right across the periodic table. As 
we go from Li across to C and on to N, O, and F, the elements steadily become more electronegative 
and the AOs lower in energy. , . 

T- 



3s 



A 



this electron 

transferred 

from 3s(Na) 

to 2p(F) 



the atomic orbitals 
are too far apart to 
combine with each 
other to form a new 
molecular orbital 



So what happens if two atoms 
whose atomic orbitals were vastly dif- 
ferent in energy, such as Na and F, were 
to combine? An electron transfers from 
sodium to fluorine and the product is 
the ionic salt, sodium fluoride, Na + F~. 

The important point is that the 
atomic orbitals are too far apart in 
energy to combine to form new mole- 
cular orbitals and no covalent bond is 
formed. The ionic bonding in NaF is 
due simply to the attraction between 
two oppositely charged ions. When 
the atomic orbitals have exactly the 
same energy, they combine to form 
new molecular orbitals, one with an 
energy lower than the AOs, the other 
with an energy higher than the AOs. 
When the AOs are very different in 
energy, electrons are transferred from 
one atom to another and ionic bond- 
ing results. When the AOs are slightly different in energy, they do combine and we need now to look 
at this situation in more detail. 



V 



2p 



Na- 



Na 



© 



F - 



sodium atom sodium ion fluoride ion fluorine atom 
both electrons in sodium fluoride end up in the fluorine's 2p orbital 



102 



4 ■ Structure of molecules 



The AOs combine to form new M Os but they do so unsymmetrically. The more electronegative 
atom, perhaps O or F, contributes more to the bonding orbital and the less electronegative element 
(carbon is the one we shall usually be interested in) contributes more to the antibonding orbital. This 
applies both to bonds and to 71 bonds so here is an idealized case. 



combine 
out-of-phase 



o> «\ 







s orbital on less 


\ 


electronegative 


\ 


element 


* 




combine 




in-phase 



o 




s orbital on more 

electronegative 

element 



molecular orbitals from elements of different electronegativity 

These three different cases where the two combining orbitals differ greatly in energy, only a little, 
or not at all are summarized below. 



Energies of AOs both the same 



AO on atom B is a little lower 
in energy than AO on atom A 



AO on atom B is a lot lower 
in energy than AO on atom A 



AO on \ 
atom A ' 



♦i yf~" 



Vb 



AO on 
atom B 



AO on 
atom A 



*Mf' 



MOs 



AO on 
atom B 



AO on 
atom A 



-H" "f 



AO on 
atom B 



Homolytic and heterolytic refer to 
the fate of the electrons when a 
bond is broken. In heterolytic 
fission one electron goes to each 
atom. In homolytic fission both 
electrons go to the same atom. 
Thus l 2 easily gives two iodine 
atoms by homolytic fission (l 2 — > 
21") while HI prefers heterolytic 
fission (HI -> H + + P). The dot in I* 
means a single unpaired electron. 



large interaction between AOs 

bonding MO much lower in energy 
than AOs 

antibonding MO is much higher in 
the energy than the AOs 

both AOs contribute equally to 
the MOs 

electrons in bonding MO are shared 
equally between the two atoms 

bond between A and B would 
classically be described as purely 
covalent 

easiest to break bond into two 
radicals (homolytic fission). 
Heterolytic fission of bond is 
possible and could give either 
A + andB"orA"andB + 



less interaction between AOs 

bonding MO is lowered only by a 
small amount relative to AO on 
atom B 

antibonding MO is raised in energy 
by only a small amount relative to 
AO on atom B 

the AO on B contributes more to the 
bonding MO and the AO on A 
electrons in bonding MO are shared 
between atoms but are associated 
more with atom B than A 
bond between A and B is covalent 
but there is also some electrostatic 
(ionic) attraction between atoms 
easiest to break bond into two ions, 
A + and B~, although it is also 
possible to give two radicals 



AOs are too far apart in energy to 

interact 

the filled orbital on the anion has the 

same energy as the AO on atom B 

the empty orbital on the cation has 
same energy as the AO on atom A 

only one AO contributes to each 'MO' 

electrons in the filled orbital are 
located only on atom B 

bond between A and B would 

classically be described as purely 

ionic 

compound already exists as ions 

A + and B - 



Heteronuclear diatomics 



103 



As an example of atomic orbitals of equal and unequal energies combining, let us consider the Jt 
bonds resulting from two carbon atoms combining and from a carbon atom combining with an oxy- 
gen atom. With the C-C Jt bond, both p orbitals have the same energy and combine to form a symmet- 
rical 71 bond. If the bonding MO (71) is occupied, the electrons are shared equally over both carbon 
atoms. Compare this with the 71 bond that results from combining an oxygen p AO with a carbon p AO. 



CC Jt*. 



,0 ^ 



p AO on 
carbon A 



p AO on 
carbon B 



CC 



*M^'> 



c 



CO jt* ,- 



c — 

a ° 



Carbon 
p AO 



\ Oxygen 
'« p AO 



CO 



71 ^ffo 



o 

C— 



Now the bonding MO (Jt) is made up with a greater contribution from the oxygen p orbital than 
from the carbon p orbital. If this MO contained electrons, there would be more electrons around the 
oxygen atom than around the carbon. This C-O Jt bond is covalent but there is also some electrosta- 
tic contribution to its bond strength. This electrostatic interaction actually makes a C-O double 
bond much stronger than a C-C double bond (bond strength for C=0, about 725-60 kjmol , for 
C=C, 600-25 kjmol : compare also a C-O single bond, 350-80 kjmol with a C-C single bond, 
340-50 kjmol - ). Because the electrons in the populated MO (71 ) are associated more with the oxy- 
gen atom than with the carbon, it is easier to break this bond heterolytically with both electrons 
moving completely on to the oxygen atom than it is to break it homolytically to get a diradical with 
one electron moving on to the carbon and one on to the oxygen atom. This will be the first chemical 
reaction we study in detail in Chapters 5 and 6. 

Other factors affecting degree of orbital interaction 

Having similar energies is not the only criterion for good inter- 
action between two atomic orbitals. It also matters how the 
orbitals overlap. We have seen that p orbitals overlap better in an 
end-on fashion (forming a bond) than they do side-on (forming 
a 7t bond). Another factor is the size of the atomic orbitals. For 
best overlap, the orbitals should be the same size — a 2p orbital 
overlaps much better with another 2p orbital than it does with a 
3p or 4p orbital. 

A third factor is the symmetry of the orbitals — two atomic orbitals must have the appropriate 
symmetry to combine. Thus a 2p x orbital cannot combine with a 2p y or 2p z orbital since they are all 
perpendicular to each other (they are orthogonal). In one case the two p orbitals have no overlap 
at all; in the other case any constructive overlap is cancelled out by equal amounts of destructive 



efficient overlap of p orbitals 

of the same size (same 
principal quantum number n) 



(+) 
R 2 C= 



(-) 
=0 




inefficient overlap 

of different size 

principal quantum 



of p orbitals 
(different 
numbers n) 




p z and p x 

these two p orbitals cannot 
combine because they are 
perpendicular to each other 




p z and p y p and s 

(side-on) 

here any constructive overlap is 

cancelled out by equal amounts 

of destructive overlap 




p and s 

(end-on) 

however, s and p orbitals 

can overlap end-on 



104 



4 ■ Structure of molecules 



overlap. Likewise, an s orbital can overlap with a p orbital only end-on. Sideways overlap leads to 
equal amounts of bonding and antibonding interactions and no overall gain in energy. 

Molecular orbitals of molecules with more than two atoms 

We now need to look at ways of combining more than two atoms at a time. For some molecules, such as 
H2S and PH3, that have all bond angles equal to 90°, the bonding should be straightforward — the p 
orbitals (which are at 90°) on the central atom simply overlap with the Is orbitals of the hydrogen atoms. 




a molecule of methane 
enclosed in a cube 




the carbon 2s AO can overlap with 
all four hydrogen Is AOs at once 




® 




H-S ©P> 







a 



the 90° angles in PH 3 and H 2 S come from the overlap of the 
hydrogen Is AO with the p AO of the phosphorus or sulfur 

But how do we account for the bond angles in water (104°) and ammonia (107°) when the only 
atomic orbitals are at 90° to each other? All the covalent compounds of elements in the row Li to Ne 
raise this difficulty. Water (H2O) and ammonia (NH3) have angles between their bonds that are 
roughly tetrahedral and methane (CH 4 ) is exactly tetrahedral but how can the atomic orbitals com- 
bine to rationalize this shape? The carbon atom has electrons only in the first and second shells, and 
the Is orbital is too low in energy to contribute to any molecular orbitals, which leaves only the 2s 
and 2p orbitals. The problem is that the 2p orbitals are at right angles to each other and methane 
does not have any 90° bonds. (So don't draw any either! Remember Chapter 2.). Let us consider 
exactly where the atoms are in methane and see if we can combine the AOs in such a way as to make 
satisfactory molecular orbitals. 

Methane has a tetrahedral structure with each C-H bond 109 pm and all the bond angles 109.5°. 
To simplify things, we shall draw a molecule of methane enclosed in a cube. It is possible to do this 
since the opposite corners of a cube describe a perfect tetrahedron. The carbon atom is at the centre 
of the cube and the four hydrogen atoms are at four of the corners. 

Now, how can the carbon's 2s and 2p atomic orbitals combine with the four hydrogen Is atomic 
orbitals? The carbon's 2s orbital can overlap with all four hydrogen Is orbitals at once with all the 
orbitals in the same phase. In more complicated systems like this, it is clearer to use a diagram of the 
AOs to see what the MO will be like. 

Each of the 2p orbitals points to opposite faces of the cube. Once more all four hydrogen Is 
orbitals can combine with each p orbital but this time the hydrogen AOs on the opposite faces of the 
cube must be differently phased. 





2 Pz 2 Py 2p x 

the hydrogen Is orbitals can overlap with the three 2p orbitals 

Again we are not going to draw these three molecular orbitals but you can see from the AO dia- 
grams what they look like. They are degenerate (that is, they have the same energy) and each orbital 
has one nodal plane (it is easiest to see in the middle diagram passing vertically down the middle of 
the cube and dividing shaded orbitals on the right from unshaded orbitals on the left). Only the 
bonding overlap between the AOs is shown but of course there is an antibonding interaction for 



Hybridization of atomic orbitals 



105 



every bonding interaction, which means there are eight MOs altogether (which is correct since there 
were eight AOs to start with). 

Organic chemists can just about understand this 'correct' MO picture of methane and theoretical 
chemists are able to construct correct MOs for very much more complex molecules than methane. 
There is experimental evidence too that these pictures are correct. Other experiments reveal that all 
four C-H bonds in methane are exactly the same and yet the MOs for methane are not all the same. 
There is no contradiction here! The molecular orbital approach tells us that there is one MO of one 
kind and three of another but the electrons in them are shared out over all five atoms. No one hydro- 
gen atom has more or less electrons than any other — they are all equivalent. Techniques that tell us 
the structure of methane do not tell us where bonds are; they simply tell us where the atoms are 
located in space — we draw in bonds connecting atoms together. Certainly the atoms form a regular 
tetrahedron but exactly where the electrons are is a different matter entirely. The classical picture of 
two atoms held together by a pair of electrons is not necessarily correct — the five atoms in methane 
are held together by electrons but these are in molecular orbitals, which spread over all the atoms. 
We are going to need the classical picture when we draw mechanisms. Methane only has one carbon 
atom — imagine what it would be like with larger compounds that can contain hundreds of carbon 
atoms! Fortunately, there is another, simpler method we can use to describe bonding that preserves 
the important points from this theory. 



Hybridization of atomic orbitals 



For most of organic chemistry, it is helpful to consider the molecule as being made up of atoms held 
together by bonds consisting of a pair of electrons. When working out the MOs for methane, we used 
the carbon 2s and all three of the 2p orbitals to combine with the hydrogen Is orbitals. Each orbital 
combined with all the hydrogen orbitals equally. Another way to consider the bonding would be to 
combine the carbon 2s and 2p orbitals first to make four new orbitals. Each of these orbitals would 
be exactly the same and be composed of one-quarter of the 2s orbital and three-quarters of one of the 
p orbitals. The new orbitals are called sp hybrid orbitals to show the proportions of the AOs in each. 
This process of mixing is called hybridization. 
Combining four atomic orbitals on the 

CXD ♦ A ♦ 3 



same atom gives the same total number of 
hybrid orbitals. Each of these has one- 
quarter s character and three-quarters p 
character. The sp orbital has a planar 
node through the nucleus like a p orbital 
but one lobe is larger than the other 
because of the extra contribution of the 2s 
orbital, which adds to one lobe but sub- 
tracts from the other. 

The four sp orbitals on one carbon 
atom point to the corners of a tetrahedron 
and methane can be formed by overlap- 
ping the large lobe of each sp orbital with 



o 



V. 



2s 2p x 2p y 

atomic orbitals all on the same atom 



2p z 



V 



J 



2s 



% -^ 



four sp 3 hybrid orbitals 

this one is from 

2s + 2p y 




2p v 



combine 
sp 3 and Is 




four sp 3 hybrid orbitals 
form a tetrahedron 



each MO orbital is the same 
and has o symmetry 



106 4 ■ Structure of molecules 



the Is orbital of a hydrogen atom. Each overlap forms an MO (2sp + Is) and we can put two elec- 
trons in each to form a C-H bond. There will of course also be an antibonding MO, o"* (2sp - Is) 

in each case, but these orbitals are empty. 
H H C-H bonds The great advantage of this method is that it can be used to build up structures of much larger 

117.8° ( / \ molecules quickly and without having to imagine that the molecule is made up from isolated atoms. 

H H ?oo b ° nd So it is easy to work out the structure of ethene (ethylene) the simplest alkene. Ethene is a planar 

molecule with bond angles close to 120°. Our approach will be to hybridize all the orbitals needed for 
(ethylene) the C-H framework and see what is left over. In this case we need three bonds from each carbon 

atom (one to make a C-C bond and two to make C-H bonds). 
Therefore we need to combine the 2s 

orbital on each carbon atom with two p 

orbitals to make the three bonds. We 

could hybridize the 2s, 2p x , and 2p y 

orbitals (that is, all the AOs in the plane) 

to form three equal sp hybrid atomic V " 

orbitals, leaving the 2p z orbital " 

unchanged. These sp hybrid orbitals will 

have one-third s character and only two- 3 x sp 2 AOs 

thirds p character. in the plane 



Q +oo ♦ X ♦ 



2s 2p x 2p y 2p z 








The three sp 2 hybrid atomic orbitals on \J unchanged 2p z 

each carbon atom can overlap with three 

other orbitals (two hydrogen Is AOs and one sp AO from the other carbon) to form three o" MOs. 
This leaves the two 2p r orbitals, one on each carbon, which combine to form the 71 MO. The skeleton 
of the molecule has five o" bonds (one C-C and four C-H) in the plane and the central 7C bond is 
formed by two 2p z orbitals above and below the plane. 

71 bond 






o bonds Q. ^B o bonds 

7C bond 

Ethyne (acetylene) has a C-C triple bond. Each carbon bonds to only two other atoms to form a 
linear CH skeleton. Only the carbon 2s and 2p x have the right symmetry to bind to only two atoms at 
once so we can hybridize these to form two sp hybrids on each carbon atom leaving the 2p y and 2p z 
to form 7t MOs with the 2p orbitals on the other carbon atom. These sp hybrids have 50% each s and 
p character and form a linear carbon skeleton. 



2s 2p x 2p y 2p z 

Y , , 

t t t 

o°o + A * 

2 x sp AOs 2p y 2p z 

two p orbitals are unchangei 



Hybridization of atomic orbitals 



107 



We could then form the MOs as shown below. Each sp hybrid AO overlaps with either a hydrogen 
Is AO or with the sp orbital from the other carbon. The two sets of p orbitals combine to give two 
mutually perpendicular K MOs. 



H 





2sp hybridized carbon atoms 
2p y and 2p z orbitals remain 




linear o bonds form skeleton 
two perpendicular jt bonds 

H — C^C — H 



Hydrocarbon skeletons are built up from tetrahedral (sp ), trigonal planar (sp ), or linear (sp) 
hybridized carbon atoms. It is not necessary for you to go through the hybridization process each 
time you want to work out the shape of a skeleton. In real life molecules are not made from their con- 
stituent atoms but from other molecules and it doesn't matter how complicated a molecule might be 
or where it comes from; it will have an easily predictable shape. All you have to do is count up the 
single bonds at each carbon atom. If there are two, that carbon atom is linear (sp hybridized), if there 
are three, that carbon atom is trigonal (sp hybridized), and, if there are four, that carbon atom is 
tetrahedral (sp hybridized). 

This hydrocarbon (hex-5-en-2-yne) has two linear sp carbon 
atoms (C2 and C3), two trigonal sp carbon atoms (C5 and C6), a 
tetrahedral sp CH 2 group in the middle of the chain (C4), and a 
tetrahedral sp methyl group (CI) at the end of the chain. We had 
no need to look at any AOs to deduce this — we needed only to 
count the bonds. 

If you had drawn the molecule more professionally as shown in the margin, you would have to 
check that you counted up to four bonds at each carbon. Of course, if you just look at the double and 
triple bonds, you will get the right answer without counting single bonds at all. Carbon atoms with 
no Jt bonds are tetrahedral (sp hybridized), those with one K bond are trigonal (sp hybridized), and 
those with two 7t bonds are linear (sp hybridized). This is essentially the VSEPRT approach with a bit 
more logic behind it. 




hex-5-en-2-yne 



All normal compounds of carbon have eight electrons in the outer shell (n = 2) of 
the carbon atom, all shared in bonds. It doesn't matter where these electrons come 
from; just fit them into the right MOs on sp, sp 



2 3 

- ,orsp atoms. 



Notice that atoms 1-4 are drawn 
in a straight line. Alkynes are 
linear — draw them like that! 




We can hybridize any atoms 

Hybridization is a property of AOs rather than specifically of carbon and, since all atoms have AOs, 
we can hybridize any atom. A tetrahedral arrangement of atoms about any central atom can be ratio- 
nalized by describing the central atom as sp hybridized. The three molecules shown here all have a 
tetrahedral structure and in each case the central atom can be considered to be sp hybridized. 

Each of these three molecules has four equivalent 6 bonds from the central tetrahedral sp atom, 
whether this is B, C, or N, and the same total number of bonding electrons — the molecules are said 
to be isoelectronic. These three elements come one after the other in the periodic table so each nucle- 
us has one more proton than the last: B has 5, C has 6, and N has 7. This is why the charge on the cen- 
tral atom varies. 

Compounds of the same three elements with only three bonds are more complicated. Borane, 
BH3, has only three pairs of bonding electrons. The central boron atom bonds to only three other 
atoms. We can therefore describe it as being sp hybridized with an empty p orbital. 

Each of the B-H bonds results from the overlap of an sp orbital with the hydrogen Is orbital. The 



,B 



H' 



10 
£"'H 
H 



borohydride anion 
H 



H' 



I 
.N„„, 



H 



ammonium cation 



/C»„„ H 

H V H 

methane 



vacant 
p orbital 



u jqiu»H trigonal 

n>u Borane 



108 



4 ■ Structure of molecules 



vacant 



sp 3 orbital tetrahedral 

/^v"'"H borane 

H X 

vacant „ ©jL,H M 8»™' 
u:i „, H — C^ methyl 



p orbital 



Q+H 



cation 



" ^ lone pair in 

\J p orbital 

N is sp 2 hybridized 



^: lone pair in 

V sp 3 orbital 

N is sp 3 hybridized^ 



V 



t'"H 



H' 



"H 



methyl anion 



H 

ammonia 



© v 



>H 



*r V 



hydronium ion 



p orbital is not needed and contains no electrons. Do not be tempted by the alternative structure 
with tetrahedral boron and an empty sp orbital. You want to populate the lowest energy orbitals for 
greatest stability and sp orbitals with their greater s character are lower in energy than sp orbitals. 
Another way to put this is that, if you have to have an empty orbital, it is better to have it of the high- 
est possible energy since it has no electrons in it and doesn't affect the stability of the molecule. 

Borane is isoelectronic with the methyl cation, CH3 . All the arguments we have just applied to 
borane also apply to Me + so it too is sp hybridized with a vacant p orbital. This will be very impor- 
tant when we discuss the reactions of carbocations in Chapter 17. 

Now what about ammonia, NH 3 ? Ammonia is not isoelectronic with borane and Me + ! As well as 
three N-H bonds, each with two electrons, the central nitrogen atom also has a lone pair of electrons. 
We have two choices: either we could hybridize the nitrogen atom sp and put the lone pair in the p 
orbital or we could hybridize the nitrogen sp and have the lone pair in an sp orbital. 

This is the opposite of the situation with borane and Me + . The extra pair of electrons does 
contribute to the energy of ammonia so it should be in the lower-energy orbital, sp , rather than 
pure p. Experimentally the H-N-H bond angles are all 107.3°. Clearly, this is much closer to the 
109.5° sp angle than to the 120° sp angle. But the bond angles are not exactly 109.5°, so ammonia 
cannot be described as pure sp hybridized. VSEPRT says the lone pair repels the bonds more than 
they repel each other. Alternatively, you could say that the orbital containing the lone pair must 
have slightly more s character while the N-H bonding orbitals must have correspondingly more p 
character. 

The methyl anion, CH3, and hydronium ion, H 3 + , are both isoelectronic with ammonia so that 
all share the same pyramidal structure. Each is approximately tetrahedral with a lone pair in an sp 
orbital. These elements follow each other in the periodic table so the change in charge occurs because 
each nucleus has one more proton than the last. VSEPRT also gives this answer. 



Shape of phosphine 

Phosphine, PH 3 , has bond angles of about 90° and there 
is no need for hybridization. The three H Is AOs can 
overlap with the three 3p orbitals of the phosphorus atom, 
which leaves the lone pair in the 3s orbital. This 'pure s' 
lone pair is less energetic and therefore less reactive than 
the sp 3 lone pair in ammonia which explains why 



ammonia is more basic than phosphine (see Chapter 8). 
In general atoms from Na to Ar are less likely to be 
hybridized than those from Li to Ne because the longer 
bonds mean the substituents are further from the central 
atom and steric interaction is less. VSEPRT does not give 
this answer. 



Double bonds to other elements 

The C=0 double bond is the most important functional group in organic chemistry. It is present in 
aldehydes, ketones, acids, esters, amides, and so on. We shall spend Chapters 5-10 discussing its 
chemistry so it is important that you understand its electronic structure. As in alkenes, the two atoms 
that make up this double bond are sp hybridized. The carbon atom uses all three sp orbitals for 
overlap with other orbitals to form bonds, but the oxygen uses only one for overlap with another 
orbital (the sp orbitals on the carbon atom) to form a O bond. However, the other two sp orbitals 
are not vacant — they contain the oxygen's two lone pairs. A p orbital from the carbon and one from 
the oxygen make up the % bond which also contains two electrons. 

C-H a bonds C-0 o bond 

I e§o • c§3 - |g^ 



lone 
pairs 



C-0 K bond 



carbon and the oxygen are both sp 



"'/»,, 



>C= 



Hybridization of atomic orbitals 



109 



The less important double bonds to nitrogen (imines) are very similar but now there is only one 
lone pair on nitrogen and a second bond to whatever substituent is on the nitrogen atom. Looking 
down on the planar structures of alkenes, imines, and ketones we see only the ends of the p orbitals 
but the rest of the structures are clearly related. 

Alkenes have a planar trigonal framework of sp carbon atoms. Each uses one sp orbital to form a 
O bond to the other carbon atom and two sp orbitals to form O bonds to the substituents (here the 
general 'R'). Two carbon p orbitals are used for a C-C Jt bond. There are no lone pairs of electrons on 
either carbon atom. 

p orbitals overlap to form a C-C ji bond 
C-C o bond between sp 2 orbitals on carbon 



trigonal planar 
sp 2 carbon 




trigonal planar 
sp 2 carbon 



bond between R and an sp orbital on carbon 

Imines have a planar trigonal framework of an sp carbon atom and an sp nitrogen atom. Each 
uses one sp orbital to form a O bond to the other atom and a p orbital to form a n bond to the other 
atom. The carbon uses two sp orbitals and the nitrogen one to form G bonds to the substituents 
(here the general 'R'). There is one lone pair of electrons on the nitrogen atom. 

p orbitals overlap to form a C-N % bond 



C-C o bond between 



trigonal planar 
sp 2 carbon 



orbitals on carbon and nitrogen 



-■ lone pair in sp orbital 
trigonal planar 




sp nitrogen 



bond between R and an sp orbital on carbon or nitrogen 



Carbonyl compounds have a planar trigonal framework of an sp carbon atom and an sp oxygen 
atom. Each uses one sp orbital to form a O bond to the other atom and a p orbital to form a K bond 
to the other atom. The carbon uses two sp orbitals to form a bonds to the substituents (here the 
general 'R'). There are two lone pairs of electrons on the oxygen atom. 

p orbitals overlap to form a C-0 n bond 
C-C c bond between sp 2 orbitals on carbon and oxygen 





R 


=(o) -*- 


■- lone pair in sp 2 orbital 


trigonal planar 
sp 2 carbon 




trigonal planar 
sp 2 oxygen 




* N 


^*- 


- lone pair in sp 2 orbital 



bond between R and an sp orbital on carbon 



Where 'R' is joined to the double bond through a carbon atom, the nature of R determines which 
orbital will be used to pair up with the sp orbital. In all the compounds shown below a saturated 
carbon atom with four bonds is joined to the double bond. The C-C single bond is a bond between 



sp carbonyl carbon 




sp J methyl group 



j ^">ni\\ "-. a bond between 
H sp 3 and sp 2 orbitals 





110 



4 ■ Structure of molecules 



an sp orbital on the ketone, imine, or alkene and an sp orbital on the substituent. It doesn't make 
any difference that the second two compounds contain rings. In all cases the black bond joins a satu- 
rated, tetrahedral, sp carbon atom to the double bond and all the black O bonds are between sp and 
sp carbons or nitrogens. 

All the other combinations are possible — here are just a few. It should be clear by now that 
bonds can form between any sort of orbitals that can point towards each other but that 7t bonds can 

form only between p orbitals. 
R 



TX 



a bond between 
two sp 2 orbitals 



o bond between 
Is and sp 2 orbitals 



o bond between 
sp and sp 2 orbitals 



Triple bonds can be formed between carbon and other elements too. The most important is the 
CN triple bond present in cyanides or nitriles. Both C and N are sp hybridized in these linear mole- 
cules, which leaves the lone pair on nitrogen in an sp orbital too. You will see (Chapter 8) how this 

affects the basicity of nitriles. 

CHN skeleton lone pair in two perpendicular 

sp orbital n bonds 



acetonitrile 



Me — C=N : 



sp J 
Me- 



) sp * 



Me- 






cr bond between c bond between 
sp 3 and sp orbitals two sp orbitals 



i All normal compounds of nitrogen have eight electrons in the outer shell (n = 2) of 
the nitrogen atom, six shared in bonds and two in a lone pair. All normal 
compounds of oxygen have eight electrons in the outer shell (n = 2) of the oxygen 
atom, four shared in bonds and four in lone pairs. It doesn't matter where these 
electrons come from; just fit them into the right MOs on sp,sp ,orsp atoms. 



Conclusion 

We have barely touched the enormous variety of molecules, but it is important that you realize at this 
point that these simple ideas of structural assembly can be applied to the most complicated mole- 
cules known. We shall use AOs and combine them into MOs to solve the structure of very small mol- 
ecules and to deduce the structures of small parts of much larger molecules. With the additional 
ideas in Chapter 7 (conjugation) you will be able to grasp the structures of any organic compound. 
From now on we shall use terms like AO and MO, 2p orbital, sp hybridization, bond, energy level, 
and populated orbital without further explanation. If you are unsure about any of them, refer back to 
this chapter for an explanation. 



Problems 

1. In the (notional and best avoided in practice) formation of NaCl 
from a sodium atom and a chlorine atom, descriptions like this 
abound in textbooks: 'an electron is transferred from the valency 
shell of the sodium atom to the valency shell of the chlorine atom'. 
What is meant, in quantum number terms, by 'valency shell'? Give a 



complete description in terms of all four quantum numbers of that 
transferred electron: (a) while it is in the sodium atom and (b) after 
it has been transferred to the chlorine atom. Why is the formation of 
NaCl by this process to be discouraged? 



Problems 



111 



2. What is the electronic structure of these species? You should 
consult a periodic table before answering. 



HP 



HS 







K 



© 



Xe 



3. What sort of bonds can be formed between s orbitals and p 
orbitals? Which will provide better overlap, Is + 2p or Is + 3p? 
Which bonds will be stronger, those between hydrogen and C, N, 
O, and F on the one hand or those between hydrogen and Si, P, S, 
and CI on the other? Within the first group, bond strength goes in 
this order: HF > OH > NH > CH. Why? 

4. Though no helium 'molecule' He2 exists, an ion HeJ does 
exist. Explain. 

5. You may be surprised to know that the molecule CH 2 , with 
divalent carbon, can exist. It is of course very unstable but it is 
known and it can have two different structures. One has an 
H-C-H bond angle of 180° and the other an angle of 120°. Suggest 
structures for these species and say which orbitals will be occupied 
by all bonding and nonbonding electrons. Which structure is like- 
ly to be more stable? 



6. Construct an MO diagram for the molecule LiH and suggest 
what type of bond it might have. 

7. Deduce the MOs for the oxygen molecule. What is the bond 
order in oxygen and where are the 2p electrons? 

8. Construct MOs for acetylene (ethyne) without hybridization. 

9. What is the shape and hybridization of each carbon atom in 
these molecules? 




CN 



H 



M 



'^ 




10. Suggest detailed structures for these molecules and predict 
their shapes. We have deliberately made noncommittal drawings 
to avoid giving away the answer to the question. Don't use these 
sorts of drawing in your answer. 

C0 2 , CH 2 =NCH 3 , CHF 3 , CH 2 =C=CH 2 , (CH 2 ) 2 



Organic reactions 



5 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• Drawing molecules realistically ch2 


• Why molecules generally don't react 


• The rest of the chapters in this book 


• Ascertaining molecular structure 


with each other! 




spectroscopically ch3 


• Why sometimes molecules do react 




• What determines molecular shape and 


with each other 




structure ch4 


• In chemical reactions electrons move 
from full to empty orbitals 

• Molecular shape and structure 
determine reactivity 

• Representing the movement of 
electrons in reactions by curly arrows 





Chemical reactions 

Most molecules are at peace with themselves. Bottles of water, or acetone (propanone, Me2C=0), or 
methyl iodide (iodomethane CH3I) can be stored for years without any change in the chemical com- 
position of the molecules inside. Yet when we add chemical reagents, say, HC1 to water, sodium 
cyanide (NaCN) to acetone, or sodium hydroxide to methyl iodide, chemical reactions occur. This 
chapter is an introduction to the reactivity of organic molecules: why they don't and why they do 
react; how we can understand reactivity in terms of charges and orbitals and the movement of elec- 
trons; how we can represent the detailed movement of electrons — the mechanism of the reaction — 
by a special device called the curly arrow. 

To understand organic chemistry you must be familiar with two languages. One, which we have 
concentrated on so far, is the structure and representation of molecules. The second is the descrip- 
tion of the reaction mechanism in terms of curly arrows and that is what we are about to start. The 
first is static and the second dynamic. The creation of new molecules is the special concern of chem- 
istry and an interest in the mechanism of chemical reactions is the special concern of organic chem- 
istry. 

Molecules react because they move. They move internally — we have seen (Chapter 3) how the 
stretching and bending of bonds can be detected by infrared spectroscopy. Whole molecules move 
continuously in space, bumping into each other, into the walls of the vessel they are in, and into the 
solvent if they are in solution. When one bond in a single molecule stretches too much it may break 
and a chemical reaction occurs. When two molecules bump into each other, they may combine with 
the formation of a new bond, and a chemical reaction occurs. We are first going to think about colli- 
sions between molecules. 

Not all collisions between molecules lead to chemical change 

All organic molecules have an outer layer of many electrons, which occupy filled orbitals, bonding 
and nonbonding. Charge-charge repulsion between these electrons ensures that all molecules repel 
each other. Reaction will occur only if the molecules are given enough energy (the activation energy 
for the reaction) for the molecules to pass the repulsion and get close enough to each other. If two 
molecules lack the required activation energy, they will simply collide, each bouncing off the elec- 
trons on the surface of the other and exchanging energy as they do so, but remain chemically 



The activation energy, also 
called the energy barrier for a 

reaction, is the minimum energy 
molecules must have if they are 
to react. A population of a given 
molecule in solution at room 
temperature has a range of 
energies. If the reaction is to 
occur, some at least must have 
an energy greater than the 
activation energy. We shall 
discuss this concept in more 
detail in Chapter 13. 



114 



5 ■ Organic reactions 



unchanged. This is rather like a collision in snooker or pool. Both balls are unchanged afterwards but 
are moving in different directions at new velocities. 



Q 



6 



on collision course 



% 



impact 



o 

after collision 



We saw why these atoms form an ionic 
compound in Chapter 4. 



We analysed the orbitals of the 
carbonyl group in Chapter 4 and 
established that the reason for the 
polarity is the greater electronegativity 
of the oxygen atom. 



electrostatic 
attraction 



0C=N 

charged 
reagent 



Charge attraction brings molecules together 

In addition to this universal repulsive force, there are also important attractive forces between mole- 
cules if they are charged. Cations (+) and anions (-) attract each other electrostatically and this may 
be enough for reaction to occur. When an alkyl chloride, RC1, reacts with sodium iodide, Nal, in ace- 
tone (propanone, Me2C=0) solution a precipitate of sodium chloride forms. Sodium ions, Na + , and 
chloride ions, CI - , ions in solution are attracted by their charges and combine to form a crystalline 
lattice of alternating cations and anions — the precipitate of crystalline sodium chloride. 

This inorganic style of attraction is rare in organic reactions. A 
more common cause of organic reactions is attraction between a 
charged reagent (cation or anion) and an organic compound that 
has a dipole. An example that we shall explore in this chapter is 
the reaction between sodium cyanide (a salt, NaCN) and a car- 
bonyl compound such as acetone. Sodium cyanide is made up 
of sodium cations, Na + , and cyanide anions, CN~, in solution. 
Acetone has a carbonyl group, a C=0 double bond, which is 
polarized because oxygen is more electronegative than carbon. 
The negative cyanide ion is attracted to the positive end of the 
carbonyl group dipole. 

It is not even necessary for the reagent to be charged. Ammonia 
also reacts with acetone and this time it is the lone pair of electrons 
— a pair of electrons not involved in bonding and concentrated on 
the nitrogen atom of the uncharged ammonia molecule — that is 
attracted to the positive end of the carbonyl dipole. 

Polarity can arise from G bonds too. The most electronegative element in the periodic table is flu- 
orine and three fluorine atoms on electropositive boron produce a partially positively charged boron 
atom by bond polarization. The negative end of the acetone dipole (the oxygen atom) is attracted 
to the boron atom in BF3. 

electrostatic 
■^ — V— ,. <- attraction 



5- 0=/d+ 



C=0 dipole 



electrostatic 
attraction 



H 



/ 



5- o^s+ -* : N— H 






C=0 dipole 



\ 
H 



lone pair 
of electrons 



F5- 



B - Fdipole ^ 5-F— B5+ 

P R V 



F5- 






C=0 dipole 

But we have not told you the whole story about BF 3 . Boron is in group 3 and thus has only six 
electrons around it in its trivalent compounds. A molecule of BF 3 is planar with an empty p orbital. 
This is the reverse of a lone pair. An empty orbital on an atom does not repel electron-rich areas of 
other molecules and so the oxygen atom of acetone is attracted electrostatically to the partial positive 
charge and one of the lone pairs on oxygen can form a bonding interaction with the empty orbital. 
We shall develop these ideas in the next section. 



Chemical reactions 



115 



So, to summarize, the presence of a dipole in a molecule represents an imbalance in the distribu- 
tion of the bonding electrons due to polarization of a bond or a % bond or to a pair of electrons or 
an empty orbital localized on one atom. When two molecules with complementary dipoles collide 
and together have the required activation energy to ensure that the collision is sufficiently energetic 
to overcome the general electronic repulsion, chemical change or reaction can occur. 

Orbital overlap brings molecules together 

Other organic reactions take place between completely uncharged molecules with no dipole 
moments. One of the old 'tests' for unsaturation was to treat the compound with bromine water. If 
the brown colour disappeared, the molecule was unsaturated. We don't use 'tests' like these any 
more (spectroscopy means we don't need to) but the reaction is still an important one. A simple 
symmetrical alkene combines with symmetrical bromine in a simple addition reaction. 

The only electrons that might be useful in the 
kind of attraction we have discussed so far are the H 
lone pair electrons on bromine. But we know from 
many experiments that electrons flow out of the 
alkene towards the bromine atom in this reac- 
tion — the reverse of what we should expect from 

electron distribution. The attraction between these molecules is not electrostatic. In fact, we know 
that reaction occurs because the bromine molecule has an empty orbital available to accept electrons. 
This is not a localized atomic orbital like that in the BF3 molecule. It is the antibonding orbital 
belonging to the Br-Br bond: the 0* orbital. There is therefore in this case an attractive interaction 
between a full orbital (the % bond) and an empty orbital (the 0* orbital of the Br-Br bond). The 
molecules are attracted to each other because this one interaction is between an empty and a full 
orbital and leads to bonding, unlike all the other repulsive interactions between filled orbitals. We 
shall develop this less obvious attraction as the chapter proceeds. 

Most organic reactions involve interactions between full and empty orbitals. Many also involve 
charge interactions, and some inorganic reactions involve nothing but charge attraction. Whatever 
the attraction between organic molecules, reactions involve electrons moving from one place to 
another. We call the details of this process the mechanism of the reaction and we need to explain 
some technical terms before discussing this. 



Br 

I 
Br 




Electron flow is the key to reactivity 

The vast majority of organic reactions are polar in nature. That is to say, electrons flow from one 
molecule to another as the reaction proceeds. The electron donor is called a nucleophile (nucleus- 
loving) while the electron acceptor is called the electrophile (electron-loving). These terms come 
from the idea of charge attraction as a dominating force in reactions. The nucleophile likes nuclei 
because they are positively charged and the electrophile likes electrons because they are negatively 
charged. Though we no longer regard reactions as controlled only by charge interactions, these 
names have stuck. 

Examples of reactions where the nucleophile is an anion and the electrophile is a cation and a new 
bond is formed simply by charge attraction leading to the combination of opposite charges include 
the reaction of sodium hydroxide with positively charged phosphorus compounds. The new bond 
between oxygen and phosphorus is formed by the donation of electrons from the nucleophile 
(hydroxide ion HO~) to the electrophile (the positively charged phosphorus atom). 



Terms such as a bond, a* orbital, it 
bond, it* orbital, lone pair, atomic and 
molecular orbital, and bonding and 
antibonding orbital, are all explained in 
Chapter 4. 



Nucleophiles do not really react 
with the nucleus but with empty 
electronic orbitals. Even so, 
electrostatic attraction (and 
repulsion) may well play a crucial 
role in determining the course of 
the process. If a molecule has a 
positive charge, it is because 
there are more protons in its 
nuclei than there are electrons 
around them. 



v 

cr ©xi 



charge 
attraction 







HO 



v 

cr I xi 

OH 



ci 

CI\|/CI 



116 



5 ■ Organic reactions 



More often, reaction occurs when electrons are transferred from a lone pair to an empty orbital as 
in the reaction between an amine and BF 3 . The amine is the nucleophile because of the lone pair of 
electrons on nitrogen and BF3 is the electrophile because of the empty p orbital on boron. 



electrophile has 
an empty orbital 





orbital 
overlap 



"F 



nucleophile has 

a lone pair 

of electrons 



Me' 



t 

Me 



Me 



©N 
Me^l^Me 
Me 



A 'dative covalent bond' is just an 
ordinary o bond whose electrons 
happen to come from one atom. Most 
bonds are formed by electron donation 
from one atom to another and a 
classification that makes it necessary 
to know the history of the molecule is 
not useful. Forget 'dative bonds' and 
stick to o bonds or jt bonds. 



The kind of bond formed in these two reactions used to be called a 'dative covalent bond' because 
both electrons in the bond were donated by the same atom. We no longer classify bonds in this 
way, but call them bonds or 71 bonds as these are the fundamentally different types of bonds in 
organic compounds. Most new bonds are formed by donation of both electrons from one atom to 
another. 

These simple charge or orbital interactions may be enough to explain simple inorganic reactions 
but we shall also be concerned with nucleophiles that supply electrons out of bonds and electrophiles 
that accept electrons into antibonding orbitals. For the moment accept that polar reactions usually 
involve electrons flowing from a nucleophile and towards an electrophile. 



• In reaction mechanisms 

• Nucleophiles donate electrons 

• Electrophiles accept electrons 

Since we are describing a dynamic process of electron movement from one molecule to another in 
this last reaction, it is natural to use some sort of arrow to represent the process. Organic chemists 
use a curved arrow (called a 'curly arrow') to show what is going on. It is a simple and eloquent sym- 
bol for chemical reactions. 

The curly arrow shows the movement 
of a pair of electrons from nitrogen into 
the gap between nitrogen and boron to 
form a new bond between those two 
atoms. This representation, what it 
means, and how it can be developed into a 
language of chemical reactions is our 
main concern in this chapter. 



0^ 



electron 
donation 



.> 



,N V 



Me^| "Me 
Me 



f 4/ f 

©N 
Me^l^Me 
Me 



the electrons 
in this new bond 
came from the 
nitrogen lone pair 



Orbital overlap controls angle of successful attack 

Electrostatic forces provide a generalized attraction between molecules in chemical reactions. In the 
reaction between chloride anions and sodium cations described above, the way in which these two 
spherical species approached one another was unimportant because the charges attracted one anoth- 
er from any angle. In most organic reactions the orbitals of the nucleophile and electrophile are 
directional and so the molecular orbitals of the reacting molecules exert important control. If a new 
bond is to be formed as the molecules collide, the orbitals of the two species must be correctly aligned 
in space. In our last example, only if the sp orbital of the lone pair on nitrogen points directly at the 
empty orbital of the BF 3 can bond formation take place. Other collisions will not lead to reaction. In 
the first frame a successful collision takes place and a bond can be formed between the orbitals. In the 
second frame are three examples of unsuccessful collisions where no orbital overlap is possible. 
There are of course many more unproductive collisions but only one productive collision. Most col- 
lisions do not lead to reaction. 



Chemical reactions 



117 




— nf F 

m/ Me 




Me^ | Me 
Me 


F F 

• *k \ / 

Mr | MeOBO 
Me | 

F 



The orbitals must also have about the right amount of energy to interact profitably. Electrons are 
to be passed from a full to an empty orbital. Full orbitals naturally tend to be of lower energy than 
empty orbitals — that is after all why they are filled! So when the electrons move into an empty orbital 
they have to go up in energy and this is part of the activation energy for the reaction. If the energy gap 
is too big, few molecules will have enough energy to climb it and reaction will be bad. The ideal 
would be to have a pair of electrons in a filled orbital on the nucleophile and an empty orbital on the 
electrophile of the same energy. There would be no gap and reaction would be easy. In real life, a 
small gap is the best we can hope for. 

Now we shall discuss a generalized example of a neutral nucleophile, Nu, with a lone pair donat- 
ing its electrons to a cationic electrophile, E, with an empty orbital. Notice the difference between the 
curly arrow for electron movement and the straight reaction arrow. Notice also that the nucleophile 
has given away electrons so it has become positively charged and that the electrophile has accepted 
electrons so it has become neutral. 



curved electron 
movement arrow 



Nu : 

lone pair on 
nucleophile 



© 



straight 
reaction arrow 



empty orbital 
on electrophile 



© 



new bond 
formed 



If we look at different possible relative energies for the lone pair orbital and the empty orbital, we 
might have equal energies, a small gap, or a large gap. Just as in Chapter 4, the horizontal lines repre- 
sent energy levels, the arrows on them represent electrons, and the vertical scale is energy with high 
energy at the top and low energy at the bottom. 



orbital 
energy 



filled orbital 


empty orbital 


of lone pair on 


on 


nucleophile 


electrophile 


If 


© 


Nu 


E 



orbitals have the same energy 




small difference in energy 
of filled and empty orbitals 



large difference in energy 
of filled and empty orbitals 



At first this picture suggests that the electrons will have to climb up to the empty orbital if it is 
higher in energy than the filled orbital. This is not quite true because, when atomic orbitals interact, 
their energies split to produce two new molecular orbitals, one above and one below the old orbitals. 
This is the basis for the static structure of molecules described in the last chapter and is also the key to 
reactivity. In these three cases this is what will happen when the orbitals interact (the new molecular 
orbitals are shown in black between the old atomic orbitals). 



These diagrams of molecular energy 
levels combining to form new bonding 
and antibonding orbitals are almost 
identical to those we used in Chapter 4 
to make molecular orbitals from atomic 
orbitals. 



118 



5 ■ Organic reactions 



1, 






new 






r^ w E ® 






new 




MOs 














MOs 




■ 1 














' ' 




■ 










orbital 
energy 




0/ \ © 

Nu 1 1 E 




/ •_ E® 

0/ / 
Nu 1 




Nu 1 


\\'l 








w 




\ilj 




new 
MOs 














orbitals have the same energj 




small difference in energy 




large difference in energy 










of filled and empty orbitals 




of filled and empty orbitals 



We saw exactly the same response 
when we combined AOs of different 
energies to make MOs in Chapter 4. 



In each case there is actually a gain in energy when the electrons from the old lone pair drop down 
into the new stable bonding molecular orbital formed by the combination of the old atomic orbitals. 
The energy gain is greatest when the two orbitals are the same and least when they are very far apart 
in energy. The other new MO is higher in energy than either of the old AOs but it does not have to be 
occupied. 

Only the highest- energy occupied orbitals of the nucleophile are likely to be similar in energy to 
only the lowest unoccupied orbitals of the electrophile. This means that the lower-lying completely 
filled bonding orbitals of the nucleophile can usually be neglected and only the highest occupied 
molecular orbital (HOMO) of the nucleophile and the lowest unoccupied molecular orbital 
(LUMO) of the electrophile are relevant. These may be of about the same energy and can then inter- 
act strongly. Orbital overlap — of both direction and energy — is therefore an important requirement 
for successful reaction between two organic molecules. 

Molecules repel each other because of their outer coatings of electrons. 
Molecules attract each other because of: 

• attraction of opposite charges 

• overlap of high-energy filled orbitals with low-energy empty orbitals 
For reaction, molecules must approach each other so that they have: 

• enough energy to overcome the repulsion 

• the right orientation to use any attraction 

We need now to look at which types of molecules are nucleophiles and which types are elec- 
trophiles. When you consider the reactivity of any molecule, this is the first question you should ask: 
is it nucleophilic or electrophilic? 

Nucleophiles donate high-energy electrons to electrophiles 

Nucleophiles are either negatively charged or neutral species with a pair of electrons in a high 
energy filled orbital that they can donate to electrophiles. The most common type of nucleo- 
phile has a nonbonding lone pair of electrons. Usually these are on a heteroatom such as O, N, S, 
or P. 



Me^l^Me 
Me 



Me^- 



Me 



water ammonia trimethylphosphine dimethylsulfide 

These four neutral molecules, ammonia, water, trimethylphosphine, and dimethylsulfide, all have 
lone pairs of electrons in sp orbitals and in each case this is the donor or nucleophilic orbital. The 
group VI atoms (O and S) have two lone pairs of equal energy. These are all nonbonding electrons 
and therefore higher in energy than any of the bonding electrons. 



Chemical reactions 



119 



H — 



H — 



hydroxide 



e 



Me 



Me- 



e 



methane thiolate 



Anions are often nucleophiles too and these are n © 

also usually on heteroatoms such as O, S, or halogen 
which may have several lone pairs of equal energy. 
The first diagram for each of our examples shows the 
basic structure and the second diagram shows all the 
lone pairs. It is not possible to allocate the negative 
charge to a particular lone pair as they are the same. 

There are a few examples of carbon nucleophiles with lone pairs of electrons, the most famous 
being the cyanide ion. Though linear cyanide has a lone pair on nitrogen and one on carbon, the 
nucleophilic atom is usually anionic carbon rather than neutral nitrogen as the sp orbital on carbon 
has a higher energy than that on the more electronegative nitrogen. Most anionic nucleophiles con- 
taining carbon have a heteroatom as the nucleophilic atom such as the anion methane thiolate 
shown above. 

Neutral carbon electrophiles usually have a 71 
bond as the nucleophilic portion of the molecule. 
When there are no lone pair electrons to supply 
high-energy nonbonding orbitals, the next best is 
the lower-energy filled 7t orbitals rather than the 
even lower-energy bonds. Simple alkenes are 
weakly nucleophilic and react with strong electrophiles such as bromine. In Chapter 20 we shall see 
that the reaction starts by donation of the 7C electrons from the alkene into the 0* orbital of the 
bromine molecule (which breaks the Br-Br bond) shown here with a curly arrow. After more steps 
the dibromoalkane is formed but the molecules are attracted by overlap between the full % orbital 
and the empty 0* orbital. 

It is possible for bonds to act as nucleophiles 
and we shall see later in this chapter that the boro- 
hydride anion, BHJ, has a nucleophilic B-H bond 
and can donate those electrons into the 71* orbital of 
a carbonyl compound breaking that bond and even- 
tually giving an alcohol as product. The first stage of the reaction has electrons from the B-H single 
bond of nucleophilic anion BH4, which lacks lone pair electrons or 7C bonds, as the nucleophile. 



©C=N 



:c=N 



n© 



© 
Q>C=N O 



cyanide ion 



sp lone pair 



sp lone pair 






H H 

\/ 



© 



H' 



H .OH 



© 



Br 



Br: 



bromide 



This point wil 
as well. 



be important in Chapter 6 



In this section you have seen lone pairs on anions and neutral molecules acting as nucleophiles 
and, more rarely, 71 bonds and even bonds able to do the same job. In each case the nucleophilic 
electrons came from the HOMO — the highest occupied molecular orbital — of the molecule. Don't 
worry if you find the curly arrows strange at the moment. They will soon be familiar. Now we need to 
look at the other side of the coin — the variety of electrophiles. 

Electrophiles have a low-energy vacant orbital 

Electrophiles are neutral or positively charged species with an empty atomic orbital (the opposite of 
a lone pair) or a low-energy antibonding orbital. The simplest electrophile is the proton, H + , a 
species without any electrons at all and a vacant Is orbital. It is so reactive that it is hardly ever found 
and almost any nucleophile will react with it. 



© 



proton 




H®^^ Nu 



H — Nu 



empty Is orbital reaction with anionic nucleophile 



Each of the nucleophiles we saw in the previous section will react with the proton and we shall 
look at two of them together. Hydroxide ion combines with a proton to give water. This reaction is 



120 



5 ■ Organic reactions 



governed by charge control. Then water itself reacts with the proton to give H30 + , the true acidic 
species in all aqueous strong acids. 



-o^\ H ® 



hydroxide as 
nucleophile 



- H )o:^h® 

H 

water as 
nucleophile 



H 
H 



We normally think of protons as acidic rather than electrophilic but an acid is just a special kind 
of electrophile. In the same way, Lewis acids such as BF 3 or AICI3 are electrophiles too. They have 
empty orbitals that are usually metallic p orbitals. We saw above how BF 3 reacted with Me 3 N. In that 
reaction BF3 was the electrophile and Me3N the nucleophile. Lewis acids such as AICI3 react violent- 
ly with water and the first step in this process is nucleophilic attack by water on the empty p orbital of 
the aluminium atom. Eventually alumina (AI2O3) is formed. 



H 



CI 



H 



p : cai o 



H 

©v 



<?./ 



CI 



H 



j n"\ 



CI 



Al 2 3 



CI 



water as empty 
nucleophile p orbital 



new 
a bond 



electrostatic 
attraction 



5- 0=(S+ - 



-ONu 



charged 
nucleophile 



C=0 dipole 



Protic and Lewis acids 

Protic acids (also known as Bransted acids) are 
electrophiles (like HCI) that can donate protons (H + ) to 
nucleophiles. They will be discussed in detail inChapter8. 
Lewis acids are also electrophiles but they donate more 



complicated cations to nucleophiles. They are usually metal 
halides such as LiCI, BF 3 , AICI 3 , SnCI 4 , and TiCI 4 . We shall 
meet them in many later chapters, particularly in Chapters 
22-8 when we discuss carbon-carbon bond formation. 



high-energy filled orbitals of the carbonyl group 



nonbonding 
lone pairs 



bonding 
orbitals 



H H 



Few organic compounds have vacant atomic orbitals and most organic electrophiles have low- 
energy antibonding orbitals. The most important are 7C* orbitals as they are lower in energy than O* 
orbitals and the carbonyl group (C=0) is the most important of these — indeed it is the most impor- 
tant functional group of all. It has a low-energy 71* orbital ready to accept electrons and also a partial 
positive charge on the carbon atom. Previously we said that charge attraction helped nucleophiles to 
find the carbon atom of the carbonyl group. 

Charge attraction is important in carbonyl reac- 
tions but so are the orbitals involved. Carbonyl com- 
pounds have a low-energy bonding Tt orbital. 
Carbonyl compounds have a dipole because in this 
filled orbital the electrons are more on electronegative 
oxygen than on carbon. The same reason (electro- 
negative oxygen) makes this an exceptionally low- 
energy orbital and the carbonyl group a very stable 
structural unit. This orbital is rarely involved in reac- 
tions. Going up the energy scale we next have two 
degenerate (equal in energy) lone pairs in nonbond- 
ing orbitals. These are the highest-energy electrons in 
the molecule (HOMO) and are the ones that react 
with electrophiles. 

When we consider the carbonyl group as an electrophile, we must look at antibonding orbitals 
too. The only one that concerns us is the relatively low-energy 71* orbital of the C=0 double bond 
(the LUMO). This orbital is biased towards the carbon to compensate for the opposite bias in the 
filled 7t orbital. How do we know this if there are no electrons in it? Simply because nucleophiles, 
whether charged or not, attack carbonyl groups at the carbon atom. They get the best overlap with 
the larger orbital component of the 7[* orbital. 




\\ 



Chemical reactions 



121 



I 

antibonding 
orbitals 


i 




nonbonding 
lone pairs 


bon 
orb 


ding 

tals 

1 






71* 



small energy gap 



=8 J! !L \JLe» 



nonbonding 
lone pair 




w 



71 



So now we can draw a mechanism for the attack of a nucleophile on the carbonyl group. 
The lone pair electrons on the nucleophile move into the 7C* orbital of the C=0 double 
bond and so break the 7t bond, though not, of course, the bond. Here is that process in 
curly arrow terms. 

The lone pair electrons on oxygen interact better with empty orbitals such as the Is of 
the proton and so carbonyl compounds are protonated on oxygen. 

The resulting cation is even more electrophilic because of the positive charge but nucle- 
ophiles still attack the carbon atom of the carbonyl group because the 7C* orbital still has 
more contribution from carbon. The positive charge is neutralized even though the nucle- 
ophile does not attack the positively charged atom. 

Even bonds can be electrophilic if the atom at one end of them is sufficiently electronegative to 
pull down the energy of the 0* orbital. Familiar examples are acids where the acidic hydrogen atom 
is joined to strongly electronegative oxygen or a halogen thus providing a dipole moment and a rela- 
tively low-energy 0* orbital. 




5- Cl- 



electrostatic 
attraction 

-HS+ -* ©B 



charged 
base 



O>c\o-(Q :Ob 



H-CI dipole 



o* orbital 
of acid 



non-bonding 
lone pair 
of base 



These two diagrams suggest two different 
ways of looking at the reaction between a base 
and an acid, but usually both interactions are 
important. Notice that an acid is just an elec- 
trophile that has an electrophilic hydrogen atom 
and a base is just a nucleophile that acts on a 
hydrogen atom. This question is explored more 
in Chapter 8. Bonds between carbon and halo- 
gen are also polarized in some cases though the 
electronegativity difference is sometimes very 
small. 

It is easy to exaggerate the importance of 
single-bond polarization. The electronegativity 
difference between H and CI is 0.9 but that 



Quick guide to 


important electronegativities 




H 










2.1 








Li 


B 


C 


N 


F 


1.0 


2.0 


2.5 


3.0 


3.5 4.0 


Mg 


Al 


Si 


P 


S CI 


1.2 


1.5 


1.8 


2.1 


2.5 3.0 

Br 

2.8 

1 
2.5 



These are Pauling 
electronegativities, calculated by 
Linus Pauling (1901-94) who 
won the chemistry Nobel prize in 
1954 and the Nobel peace prize 
in 1983 and from whose ideas 
most modern concepts of the 
chemical bond are derived. Born 
in Portland, Oregon, he worked at 
'CalTech' (the California Institute 
of Technology at Pasadena) and 
had exceptionally wide-ranging 
interests in crystallography, 
inorganic chemistry, protein 
structure, quantum mechanics, 
nuclear disarmament, politics, 
and taking vitamin C to prevent 
the common cold. 



122 



5 ■ Organic reactions 



between C and Br only 0.3 while the C-I bond is not polarized at all. When carbon-halogen bonds 
act as electrophiles, polarity hardly matters but a relatively low-energy 0* orbital is vitally important. 
The bond strength is also important in these reactions too as we shall see. 

Some bonds are electrophilic even though they have no dipole at all. The halogens such as 
bromine (Br2) are examples. Bromine is strongly electrophilic because it has a very weak Br-Br 
bond. Symmetrical bonds have the energies of the orbital and the 0* orbital roughly evenly dis- 
tributed about the nonbonding level. A weak symmetrical bond means a small energy gap while a 
strong symmetrical bond means a large energy gap. Bromine is electrophilic but carbon-carbon 
bonds are not. Reverting to the language of Chapter 4, we could say that the hydrocarbon framework 
is made up of strong C-C bonds with low-energy populated and high-energy unpopulated orbitals, 
while the functional groups react because they have low LUMOs or high HOMOs. 



small 

energy 

gap 

small 

energy 

gap 

weak 

bond 



Br — Br 



Br — Br 



Br — Br 




^ r six degenerate 

nonbonding 

lone pairs 

w 



nonbonding 
electrons of 
nucleophile 



large 

energy 

gap 

strong 

bond 



H3C CH3 



Notice how putting charges in circles 
(Chapter 2) helps here. There is no 
problem in distinguishing the charge on 
sulfur (in a ring) with the plus sign (not 
in a ring) linking the two products of the 
reaction. 



An example would be the rapid reaction between a sulfide and bromine. No reaction at all occurs 
between a sulfide and ethane or any other simple C-C bond. Lone pair electrons are donated from 
sulfur into the Br-Br 0* orbital, which makes a new bond between S and Br and breaks the old 
Br-Br bond. 



Sr-^-Bt ( : 



Me 

/ 



Me 



e 



Br 



Me 

-S© 
\ 
Me 



Summary: interaction between HOMO and LUMO leads to reaction 

Organic reactions occur when the HOMO of a nucleophile overlaps with the LUMO of the elec- 
trophile to form a new bond. The two electrons in the HOMO slot into the empty LUMO. The react- 
ing species may be initially drawn together by electrostatic interaction of charges or dipoles but this is 
not necessary. Thus at this simplest of levels molecular recognition is required for reaction. The two 
components of a reaction must be matched in terms of both charge-charge attraction and the energy 
and orientation of the orbitals involved. 

Nucleophiles may donate electrons (in order of preference) from a lone pair, a 7t bond, or even a 
bond and electrophiles may accept electrons (again in order of preference) into an empty orbital or 
into the antibonding orbital of a 7t bond (71* orbital) or even a bond (0* orbital). These antibond- 
ing orbitals are of low enough energy to react if the bond is very polarized by a large electronegativity 
difference between the atoms at its ends or, even for unpolarized bonds, if the bond is weak. 

The hydrocarbon framework of organic molecules is unreactive. Functional groups such as NH 2 
and OH are nucleophilic because they have nonbonding lone pairs. Carbonyl compounds and alkyl 
halides are electrophilic functional groups because they have low-energy LUMOs (7t* for C=0 and 
0* for C-X, respectively). 



Organic chemists use curly arrows to represent reaction mechanisms 



123 



Organic chemists use curly arrows to represent reaction 
mechanisms 

You have seen several examples of curly arrows so far and you may already have a general idea of 
what they mean. The representation of organic reaction mechanisms by this means is so important 
that we must now make quite sure that you do indeed understand exactly what is meant by a curly 
arrow, how to use it, and how to interpret mechanistic diagrams as well as structural diagrams. 

A curly arrow represents the actual movement of a pair of electrons from a filled orbital into an 
empty orbital. You can think of the curly arrow as representing a pair of electrons thrown, like a 
climber's grappling hook, across from where he is standing to where he wants to go. In the simplest 
cases, the result of this movement is to form a bond between a nucleophile and an electrophile. Here 
are two examples we have already seen in which lone pair electrons are transferred to empty atomic 
orbitals. 



H— ^H u 



H 



hydroxide ion 
as nucleophile 



empty 
Is orbital 



new 
o bond 



I CI 

vr>Jo 



water as empty 
nucleophile p orbital 



/ 



@ /°T A V cl 

H CI 

new 
a bond 



Note the exact position of the curly arrow as the value of this representation lies in the precision 
and uniformity of its use. The arrow always starts with its tail on the source of the moving electrons, 
representing the filled orbital involved in the reaction. The head of the arrow indicates the final des- 
tination of the pair of electrons — the new bond between oxygen and hydrogen or oxygen and alu- 
minium in these examples. As we are forming a new bond, the head of the arrow should be drawn to 
a point on the line between the two atoms. 

When the nucleophile attacks an antibonding orbital, such as the weak Br-Br bond we have just 
been discussing, we shall need two arrows, one to make the new bond and one to break the old. 



Br-<-Br t : 



Me 

/ 
> 

\ 

Me 



O 



-*- Br 



Me 

-S© 
\ 
Me 



The bond-making arrow is the same as before but the bond-breaking arrow is new. This arrow 
shows that the two electrons in the bond move to one end (a bromine atom) and turn it into an anion. 
This arrow should start in the centre of the bond and its head should rest on the atom (Br in this case) 
at the end of the bond. Another example would be the attack of a base on the strong acid HBr. 



\ 



Br 



© / 
— N— 

\ 



Some chemists prefer to place 
this point halfway between the 
atoms but we consider that the 
representation is clearer and 
more informative if the arrowhead 
is closer to the atom to which the 
new bond is forming. For these 
examples the difference is 
minimal and either method is 
completely clear but in more 
complex situations, our method 
prevents ambiguity as we shall 
see later. We shall adopt this 
convention throughout this book. 



It is not important how much curvature you put into the arrows or whether they are above or 
below the gaps of the bonds, both on the same side, or on opposite sides so long as they begin and 
end in the right places. All that matters is that someone who reads your arrows should be able to 
deduce exactly what is happening in the reaction from your arrows. We could have drawn the 
ammonia/HBr reaction like this if we had wished. 



H 
\ 
H — N 
/ 

H 



H-^Br 



H 

\© 
— N — H 
/ 
H 



© 



Br 



Charge is conserved in each step of a reaction 

In all these examples we have reacted neutral molecules together to form charged species. Because 
the starting materials had no overall charge, neither must the products. If we start with neutral 
molecules and make a cation, we must make an anion too. Charge cannot be created or destroyed. If 



124 



5 ■ Organic reactions 



our starting materials have an overall charge — plus or minus — then the same charge must appear in 
the products. 



H H 

starting materials have 
overall positive charge 



V 



H 

©/ 
H — N — H 
\ 
H 



products must also have 
overall positive charge 



When it is a 71 bond that is being broken rather than a bond, only the 7t bond is broken and the 
O bond should be left in place. This is what commonly happens when an electrophilic carbonyl 
group is attacked by a nucleophile. Just as in the breaking of a bond, start the arrow in the middle 
of the Jt bond and end by putting the arrowhead on the more electronegative atom, in this case oxy- 
gen rather than carbon. 




"X- 



■ C-0 a bond 
remains 



it bond is broken 



In this case the starting materials had an overall negative charge and this is preserved as the 
oxyanion in the product. The charge disappears from the hydroxide ion because it is now sharing a 
pair of electrons with what was the carbonyl carbon atom and a charge appears on what was the car- 
bonyl oxygen atom because it now has both of the electrons in the old 71 bond. 

Electrons can be donated from 7t bonds and from O bonds too. The reaction of an alkene with 
HBr is a simple example of a C-C 7C bond as nucleophile. The first arrow (on the nucleophile) starts 
in the middle of the % bond and goes into the gap between one of the carbon atoms and the hydrogen 
atom of HBr. The second arrow (on the electrophile) takes the electrons out of the H-Br bond and 
puts them on to the bromine atom to make bromide ion. This sort of reaction make us place alkenes 
among the functional groups as well as as part of the framework of organic molecules. 



X=c\ H ^ 



Br 






g 



Br 



We have drawn in the hydrogen atom 
that was part of HBr. It is not necessary 
to do this but you may wish to show 
what has happened to one particular 
hydrogen atom among many in a 
reaction mechanism and this is another 
instance of ignoring, for a good reason, 
one of the guidelines from Chapter 2! 



Notice that it was important to draw the two reagents in the right orientation since both are 
unsymmetrical and we want our arrow to show which end of the alkene reacts with which end of 
HBr. If we had drawn them differently we should have had trouble drawing the mechanism. Here is a 
less satisfactory representation. 




H-^Br 








/'© 



Br 



If you find yourself making a drawing like this, it is worth having another go to see if you can be 
clearer. Drawing mechanisms is often rather experimental — try something and see how it looks: if it 
is unclear, try again. One way to avoid this particular problem is to draw an atom-specific curly arrow 
passing through the atom that reacts. Something like this will do. 



sSr 



H-^Bi 



Br 




Br 




Br 



This reaction does not, in fact, stop here as the two ions produced (charge conserva- 
tion requires one cation and one anion in this first reaction) now react with each other to 
form the product of the reaction. This reaction is pretty obvious as the anion is the nucle- 
ophile and the cation, with its empty p orbital, is the electrophile. 



Organic chemists use curly arrows to represent reaction mechanisms 125 

The reaction that occurs between the alkene and HBr occurs in two stages — the formation of the 
ions and their combination. Many reactions are like this and we call the two stages steps so that we 
talk about 'the first step' and 'the second step', and we call the ions intermediates because they are 
formed in one step and disappear in the next. We shall discuss these intermediates in several later 
chapters (for example, 17 and 19). 

When O bonds act as nucleophiles, the electrons also have to go to one end of the O bond as they 
form a new bond to the electrophile. We can return to an earlier example, the reaction of sodium 
borohydride (NaBH^ with a carbonyl compound, and complete the mechanism. In this example, 
one of the atoms (the hydrogen atom) moves 
away from the rest of the BH4 anion and H H 0-*v H H ^ q© 

becomes bonded to the carbonyl compound. y^~-C "^ ill *~ s° + 

The LUMO of the electrophile is, of course, H © H x"\ H 

the 71* orbital of the C=0 double bond. 

The arrow on the nucleophile should 
again start in the middle of the bond that 

breaks and show which atom (the black H in „ „ u „ ^ 

/ &*~} / H 

this case) is transferred to the electrophile. \ / |Ly \ / \ / 

The second arrow we have seen before. Here H^®X^tC /"^\ W^ /^\ 

again you could use an atom-specific arrow 
to make it clear that the electrons in the O 
bond act as a nucleophile through the hydro- 
gen and not through the boron atom. 

This reaction also occurs in two steps and &~\ (^^\ 

. . ,. , H ,CT f H ^H H OH p. 

the oxyamon is an intermediate, not a prod- \ / \ / + ho 



X — X 



HO 

uct. The reaction is normally carried out in 

water and the oxyanion reacts with water by 
proton transfer. 

We shall discuss this reaction, the reduction of carbonyl compounds by NaBH.4, in detail in 
Chapter 6. 

The decomposition of molecules 

So far we have described reactions involving the combination of one molecule with another. Many 
reactions are not like this but involve the spontaneous decomposition of one molecule by itself 
without any assistance from any other molecule. In these reactions there is no electrophile or 
nucleophile. The usual style of reaction consists of a weak, often polarized O bond breaking to give 
two new molecules or ions. The dissociation of a strong acid HX is a simple example. 

In organic chemistry spontaneous dissociation of diazonium salts, compounds H-£ — X »- H + Br 

containing the Nj group, occurs very easily because one of the products, nitrogen 

gas ('dinitrogen') is very stable. It does not much matter what R is (alkyl or aryl); this R— ; — N=N »- R® + N=N 

reaction happens spontaneously at room temperature. *-? 

This is not, of course, the end of the reaction as R + is very reactive and we shall see the sort of 
things it can do in Chapters 17 and 19. More commonly, some sort of catalysis is involved in decom- 
position reactions. An important example is the decomposition of tertiary alcohols in acid solution. 
The carbon-oxygen bond of the alcohol does not break by itself but, after the oxygen atom has been 
protonated by the acid, decomposition occurs. 



-^N>h\®h 





H,0 



© 



This two-step mechanism is not finished because the positive ion (one particular example of R + ) 
reacts further (Chapter 17). In the decomposition step the positive charge on the oxygen atom as 
well as the fact that the other product is water helps to break the strong C-O bond. In these three 



126 



5 ■ Organic reactions 



examples, the functional group that makes off with the electrons of the old bond (X, Nj , and 
OHj ) is called the leaving group, and we shall be using this term throughout the book. The sponta- 
neous decomposition of molecules is one of the clearest demonstrations that curly arrows mean the 
movement of two electrons. Chemical reactions are dynamic processes, molecules really do move, 
and electrons really do leave one atomic or molecular orbital to form another. 

These three examples all have the leaving group taking both electrons from the old bond. This 
type of decomposition is sometimes called heterolytic fission or simply heterolysis and is the most 
common in organic chemistry. There is another way that a bond can break. Rather than a pair of 
electrons moving to one of the atoms, one electron can go in either direction. This is known as 
homolytic fission as two species of the same charge (neutral) will be formed. It normally occurs when 

similar or indeed identical atoms are at each end of the 

bond to be broken. Both fragments have an unpaired elec- 

vJ tron and are known as radicals. This type of reaction occurs 

when bromine gas is subjected to sunlight. 



Each bromine radical has an 
unpaired electron in an atomic 
orbital. 



Br 



-j— Br 



-»- Br 



Br 



Don't be alarmed — these mechanisms 
will all be discussed in full later in the 
book, this particular one in Chapter 10. 



The weak Br-Br bond breaks to form two bromine radicals. This can be represented by two sin- 
gle-headed curly arrows, fish hooks, to indicate that only one electron is moving. This is virtually all 
you will see of this special type of curly arrow until we consider the reactions of radicals in more 
detail (Chapter 39). When you meet a new reaction you should assume that it is an ionic reaction 
and use two-electron arrows unless you have a good reason to suppose otherwise. 

Curly arrows also show movement of electrons within molecules 

So far all the mechanisms we have drawn have used only one or two arrows in each step. In fact, there 

is no limit to the number of arrows that might be involved and we need to look at some mechanisms 

v „ with three arrows. The third arrow in such mechanisms 

s — x V==0 »- \ n® usually represents movement of electrons inside of the 

Nu / / reacting molecules. Some pages back we drew out the 

addition of a nucleophile to a carbonyl compound. 

This is a two-arrow mechanism but, if we lengthen the structure of the carbonyl compound by 
adding a double bond in the right position, we can add the nucleophile to a different position in the 
molecule by moving electrons within the molecule using a third arrow. 

The first arrow from the nucleophile 
makes a new bond and the last breaks the 
carbonyl 71 bond. The middle arrow just 
moves the C-C % bond along the molecule. If 
you inspect the product you will see that its 
structure follows precisely from the arrows. The middle arrow starts in the middle of a % bond and ends in 
the middle of a bond. All it does is to move the % electrons along the molecule. It turns the old % bond 
into a bond and the old bond into a % bond. We shall discuss this sort of reaction in Chapter 10. 

In some mechanisms there is a second step in the mechanism and both are three-arrow processes. 
Here is the first step in such a mechanism. See if you can understand each arrow before reading the 
explanation in the next paragraph. 



& 
Nu 



^A 





h 



.© 




hot *► Nr 

The arrow from the hydroxide ion removes a proton from the molecule making a new O-H bond 
in a molecule of water. The middle arrow moves the electrons of a C-H bond into a C-C bond mak- 
ing it into a Jt bond and the third arrow polarizes the carbonyl 71 bond leaving an oxyanion as the 
product. Charge is conserved — an anion gives an anion. In fact this 'product' is only an intermediate 
and the second step also involves three arrows. 



Drawing your own mechanisms with curly arrows 127 

Starting from the oxyanion, the first arrow re-forms the carbonyl group, the £> . 

middle arrow moves a K bond along the molecule, and the third arrow breaks a WJ © 

C-O bond releasing hydroxide ion as one of the products of the reaction. We HO^^^^^^^ 
shall meet this sort of reaction in detail later (Chapters 19 and 27). 

Mostly for entertainment value we shall end this section with a mechanism involving no fewer 
than eight arrows. See if you can draw the product of this reaction without looking at the result. 

The first arrow forms a new C-S bond and the last arrow breaks a C-Br bond but all the rest 
just move % bonds along the molecule. The product is therefore: 
MeS 



We shall not be discussing this reaction anywhere in the book! We have included it just to con- 
vince you that, once you understand the principle of curly arrows, you can understand even very 
complicated mechanisms quite easily. At this stage we can summarize the things you have learned 
about interpreting a mechanism drawn by someone else. 

Summary: what do curly arrows mean? 

• A curly arrow shows the movement of a pair of electrons 

• The tail of the arrow shows the source of the electron pair, which will be a filled orbital (HOMO) 

• such as a lone pair or a % bond or a bond 

• The head of the arrow indicates the ultimate destination of the electron pair which will either be: 

• an electronegative atom that can support a negative charge (a leaving group) 

• or an empty orbital (LUMO) when a new bond will be formed 

• or an antibonding orbital (jt* or 0*) when that bond will break 

• Overall charge is always conserved in a reaction. Check that your product obeys this rule 

Now would be a good time to do Problems 1 and 2 at the end of the chapter, which will give you 
practice in the interpretation of mechanisms. 

Drawing your own mechanisms with curly arrows 

Curly arrows must be drawn carefully! The main thing you need to remember is that curly arrows 
must start where there is a pair of electrons and end somewhere where you can leave a pair of elec- 
trons without drawing an absurd structure. That sounds very simple — and it is — but you need some 
practice to see what it means in detail in different circumstances. Let us look at the implications with 
a reaction whose products are given: the reaction of © q 

triphenylphosphine with methyl iodide. + P " 3P *~ 3 

First observe what has happened: a new bond has been formed between the phosphorus atom and 
the methyl group and the carbon-iodine bond has been broken. Arrows represent movement of elec- 
tron pairs not atoms so the reactants must be drawn within bonding distance before the mechanism 
can be drawn. This is analogous to the requirement that molecules must collide before they can react. 
First draw the two molecules so that the atoms that form the new bond (P and C) are near each other Ph 3 P CH3 
and draw out the bonds that are involved (that is, replace 'Mel' with a proper chemical structure). 

Now ask: which is the electrophile and which the nucleophile (and why)? The phosphorus atom 
has a lone pair and the carbon atom does not so Ph 3 P must be the nucleophile and the C-I bond of 
Mel must be the electrophile. All that remains ^^ f\ 

is to draw the arrows. ph 3 p : ' CH 3 ' *~ Ph 3 p CH 3 + I" 

Admittedly, that was quite an easy mechanism to draw but you should still be pleased if you suc- 
ceeded at your first try. 



128 



5 ■ Organic reactions 



Warning! Eight electrons is the maximum for B, C, N, or O 

We now ought to spell out one thing that we have never stated but rather assumed. Most organic 
atoms, if they are not positively charged, have their full complement of electrons (two in the case of 
hydrogen, eight in the cases of carbon, nitrogen, and oxygen) and so, if you make a new bond to one 
of those elements, you must also break an existing bond. Suppose you just 'added' PI13P to Mel in this 
last example without breaking the C-I bond: what would happen? 



,r\ 



Ph 3 P: ^CHa 

wrong mechanism 



^^ 



H 

H H 

impossible structure 
carbon has five bonds 



This structure must be wrong because carbon cannot have five bonds — if it did it would have ten 
electrons in the 2s and the 2p orbitals. As there are only four of those (2s, 2\)x, 2p„ and 2p z ) and they 
can have only two electrons each, eight electrons is the maximum and that means that four bonds is 
the maximum. 



# If you make a new bond to uncharged H, C, N, or O you must also break one of the 
existing bonds in the same step. 

There is a nasty trap when a charged atom has its full complement of electrons. Since BH4 and 
NH4 are isoelectronic with methane and have four o" bonds and hence eight electrons, no new bonds 
can be made to B or N. The following attractive mechanisms are impossible because boron has no 
lone pair in BH4 and nitrogen has no empty orbital in NH4 . 



© 



H H 



H 



H H 



-x 



"-X- Y 

H H 




impossible reaction 



impossible structure impossible structure impossible structure 
boron has five bonds carbon has five bonds nitrogen has five bonds 



impossible reaction 



Reactions with BH4 always involve the loss of H and a pair of electrons using the BH bond as 
nucleophile and reactions with NH4 always involve the loss of H without a pair of electrons using the 
NH bond as electrophile. 



H H 



,© 



H' "H T X" 

correct mechanism 



H 



H H 
\/ 
^B + HX 




Similarly, nucleophiles do not attack species like H^O" 1- at oxygen, even though it is the oxygen 
atom that carries the positive charge. Reaction occurs at one of the protons, which also neutralizes 
the positive charge. Or, to put it another way, H30 + is an acid (electrophilic at hydrogen) and not 
electrophilic at oxygen. 



Y H 






impossible structure 
oxygen has four bonds 



Y^©l 



impossible reaction 



H ^H >03. 



correct mechanism 



H 

I 
JQ + HY 



■ © 



^OH 



Try a simple example: primary alcohols can be con- 
verted into symmetrical ethers in acid solution. Suggest 
a mechanism for this acid-catalysed conversion of one 
functional group into another. 



Drawing your own mechanisms with curly arrows 



129 




The reaction must start by the protonation of something and the 
only candidate is the oxygen atom as it alone has lone pair electrons. 
This gives us a typical oxonium ion with three bonds to oxygen and a 
full outer shell of eight electrons. 

To make the ether a second molecule of alcohol must be added but 
we must not now be tempted to attack the positively charged oxygen 
atom with the nucleophilic OH group. The second molecule could 
attack a proton, but that would just make the same molecules. Instead it 
must attack at carbon expelling a molecule of water as a leaving group 
and creating a new oxonium ion. 

Finally, the loss of the proton from the new oxonium ion gives the ether. Though this is a three- 
step mechanism, two of the steps are just proton transfers in acidic solution and the only interesting 
step is the middle one. Here is the whole mechanism. 




J 




c 



oxonium ion 




+ H 2 



NaCN 



H,0 



Drawing a two-step mechanism: cyanohydrin formation 

Now what about this slightly more complicated example? Sodium 
cyanide is added to a simple aldehyde in aqueous solution. The product 
is a cyanohydrin and we shall discuss this chemistry in Chapter 6. 

This reaction is presented in a style with which you will become familiar. The organic starting 
material is written first and then the reagent over the reaction arrow and the solvent under it. We 
must decide what happens. NaCN is an ionic solid so the true ^_ 

reagent must be cyanide ion. As it is an anion, it must be the — II 



nucleophile and the carbonyl group must be the electrophile. Let 
us try a mechanism. 

This is a good mechanism but it doesn't quite produce the product. There must be a second step 
in which the oxyanion picks up a proton from somewhere. The only source of protons is the solvent, 
water, so we can write: 




h^hOo 



.CN 



H 




CN 



H 



s H -- - -^ -. H ^ 

This is the complete mechanism and we can even make a prediction about the reaction conditions 
from it. The second step needs a proton and water is not a very good proton donor. A weak acid as 
catalyst would help. 

Now for a real test: can you draw a mechanism for this reaction? 



base 



OH 



You might well protest that you don't know anything about the chemistry of three-membered 
rings or of either of the functional groups, SH and cyclic ether. Be that as it may, you can still draw a 
mechanism for the reaction. It is important that you are prepared to try your hand at mechanisms 
for new reactions as you can learn a lot this way. Ask first of all: which bonds have been formed and 
which broken? Clearly the S-H bond has been broken and a new S-C bond formed. The three-mem- 
bered ring has gone by the cleavage of one of the C-O bonds. The main chain of carbon atoms is 
unchanged. We might show these ideas in some way such as this. 



HO, CN 
~H 



0, 



CN 
H 



new bond formed 
between these atoms 




this bond 
is broken 



this bond is broken 



130 5 ■ Organic reactions 



Now you could continue in many ways. You might say 'what breaks the SH bond?' This must be the 
role of the base as a base removes protons. You might realize that the reaction cannot happen while the 



A 



sulfur atom is so far away from the three- 

■ ■ ./\ /\ /I base © 

membered ring (no chance of a collision) HS ^^ ^■^ ' 

and redraw the molecule so that the reac 
tion can happen. 

Now draw the mechanism. It is easy once you have done the preparatory thinking. The sulfur 
anion must be the nucleophile so the C-O bond in the three-membered ring must be the elec- 
trophile. Here goes! 

That is not quite the product so we 
must add a proton to the oxyanion. ^ 
Where can the proton come from? 
It must be the proton originally re- 
moved by the base as there is no other. 5 /-n 
We can write B for the base and hence \ y^ ^H-r-B® »~ ( "V ^OH 



t> 



0© 



BH + for the base after it has captured a \ / ^ 

proton. 

Your mechanism probably didn't look as neat as the printed version but, if you got it roughly 
right, you should be proud. This is a three-step mechanism involving chemistry unknown to you and 
yet you could draw a mechanism for it. Are you using coloured arrows, by the way? We are using 
black arrows on red diagrams but the only point of that is to make the arrows stand out. We suggest 
you use any colour for your arrows that contrasts with your normal ink. 

Decide on a 'push' or a 'pull' mechanism 

In one step of a reaction mechanism electrons flow from a site rich in electrons to an electron-defi- 
cient site. When you draw a mechanism you must make sure that the electrons flow in one direction 
only and neither meet at a point nor diverge from a point. One way to do this is to decide whether the 
mechanism is 'pushed' by, say, a lone pair or an anion or whether it is 'pulled' by, say, a cation, an 
empty orbital, or by the breaking of a reactive weak 71 bond or bond. This is not just a device either. 
Extremely reactive molecules, such as fluorine gas, F2, react with almost anything — in this case 
because of the very electrophilic F-F bond (low energy F-F 0* orbital). Reactions of F2 are 'pulled' 
by the breaking of the F-F bond. The nearest thing in organic chemistry is probably the reactions of 
carbon cations such as those formed ^ s~^ 

by the decomposition of diazonium R— — N^N *- R Nu >~ R — Nu 

salts. 

In the first step the electrons of the bond are pulled away by the positive charge and the very sta- 
ble leaving group, N 2 . In the second step lone pair electrons are pulled into the very reactive cation 
by the nonbonding empty orbital on carbon. Even very weak nucleophiles such as water will react 
with such cations as a real example shows. 





H 



©^H 




In all our previous examples we have drawn the first arrow from the nucleophile, anion, lone pair, 
or whatever and pushed the electrons along the chain of arrows. This is a natural thing to do; indeed 
the skill of drawing mechanisms is sometimes derisively referred to as 'electron pushing', but some 
mechanisms are more easily understood as 'electron pulling'. In general, if a cation, an acid, or a 
Lewis acid is a reagent or a catalyst, the reaction is probably pulled. If an anion or a base is involved as 
a reagent, the reaction is probably pushed. In any case it isn't so important which approach you adopt 
as that you should do one or the other and not muddle them up. 



Drawing your own mechanisms with curly arrows 



131 



^^ 



HBr 



number of protons on each carbon atom 
in starting material and product 



A more interesting example of a pull mecha- 
nism is the reaction of isoprene (2-methylbutadi- 
ene) with HBr. The product is an unsaturated 
alkyl bromide (a bromoalkene). 

What has happened? HBr has clearly added to the diene while one of the double bonds has van- 
ished. However, the remaining double bond, 
whichever it is, has moved to a new position in the 
middle of the molecule. So how do we start? HBr 
is a strong acid so the reaction must begin with the 
protonation of some atom in the diene by HBr. 
Which one? If you examine the product you will 
see that one atom has an extra hydrogen and this 
must be where protonation occurs. 

The only change is at the left-hand end of the molecule where there is an extra proton. We must 
add the proton of HBr to that atom. The highest-energy orbital at that atom is the rather unreactive 



* 



.Wvi 



HBr 




I 



A T* Br 



2 12 



3 12 



Br 



n* r^^s 




© 



alkene 7t bond so we must use that as the 
nucleophile, though the electrons are real- 
ly being pulled out of the 7t bond by reac- 
tive HBr. 

It is not necessary to draw in that hydrogen atom in the product of this step. It is, of course, 
necessary to put the positive charge on the carbon atom in the middle that has lost electrons. Now 
we can add bromide ion (the other product of the first step) to this cation but not where we have 
written the plus charge as that will not give us the right product. We must move the remaining 

double bond along the molecule as we 

add the bromide ion. This too is a 'pulled' 

reaction as the unstable plus charge on 

carbon pulls electrons towards itself. 

So this is a two-step reaction and the driving force for the two steps is a strongly acidic electro- 

phile in the first and a strongly electrophilic cation at carbon in the second. Here is the full 

mechanism. 



/<k>^ 



n 



e 



Br 



b Cl 



r^y^^ ^ /40* ^°Br 



Now we can summarize the extra points we have made in this section as a series of guidelines. 

Extra guidelines for writing your own mechanisms 

• Decide on the structure of any ambiguous reagents, for example, salt or a covalent compound? 

• Decide which is the nucleophilic and which the electrophilic atom 

• Decide whether to think in a push or a pull manner 

• Mark lone pairs on the nucleophilic atom 

• Draw the molecule(s) in a spatial arrangement that makes reaction possible 

• Curly arrows always move in the same direction. They never meet head on! 

• If you make a new bond to H, C, N, or O you must also break one of the existing bonds in the 
same step 

• Draw your arrows in colour to make them stand out 

• Mark charges clearly on reactants and intermediates 

• Make sure that overall charge is conserved in your mechanism 

We have only given you a preliminary trial run as a learner driver of curly arrows in this section. 
The way forward is practice, practice, practice. 



132 5 ■ Organic reactions 



Curly arrows are vital for learning organic chemistry 

Curly arrows can be used to explain the interaction between the structure of reactants and products 
and their reactivity in the vast majority of organic reactions, regardless of their complexity. When 
used correctly they can even be used to predict possible outcomes of unknown processes and hence 
to design new synthetic reactions. They are thus a powerful tool for understanding and developing 
organic chemistry and it is vital that you become proficient in their use. They are the dynamic lan- 
guage of organic reaction mechanisms and they will appear in every chapter of the book from now 
on. 

Another equally important reason for mastering curly arrows now, before you start the systemat- 
ic study of different types of reactions, is that the vast number of 'different reactions' turn out not to 
be so different after all. Most organic reactions are ionic; they therefore all involve nucleophiles and 
electrophiles and two-electron arrows. There are relatively few types of organic electrophiles and 
nucleophiles and they are involved in all the 'different' reactions. If you understand and can draw 
mechanisms, the similarity between seemingly unrelated reactions will become immediately appar- 
ent and thus the number of distinct reaction types is dramatically reduced. 

Drawing curly arrow mechanisms is a bit like riding a bike. Before you've mastered the skill, you 
keep falling off. Once you've mastered the skill, it seems so straightforward that you wonder how you 
ever did without it. You still come across busy streets and complex traffic junctions, but the basic 
skill remains the same. 

If you still feel that drawing mechanisms for yourself is difficult, this stage-by-stage guide may 
help you. Once you've got the idea, you probably won't need to follow it through in detail. 

A guide to drawing mechanisms with curly arrows 

1 Draw out the reagents as clear structures following the guidelines in Chapter 2. Check that you 
understand what the reagents and the solvent are under the conditions of the reaction, for 
example, if the reaction is in a base, will one of the compounds exist as an anion? 

2 Inspect the starting materials and the products and assess what has happened in the reaction. 
What new bonds have been formed? What bonds have been broken? Has anything been added or 
removed? Have any bonds moved around the molecule? 

3 Identify the nucleophilic centres in all the reactant molecules and decide which is the most 
nucleophilic. Then identify the electrophiles present and again decide which is the most 
electrophilic 

4 If the combination of these two centres appears to lead to the product, draw the reactants, 
complete with charges, so as to position the nucleophilic and electrophilic centres within 
bonding distance ensuring that the angle of attack of the nucleophile is more or less consistent 
with the orbitals involved 

5 Draw a curly arrow from the nucleophile to the electrophile. It must start on the filled orbital or 
negative charge (show this clearly by just touching the bond or charge) and finish on the empty 
orbital (show this clearly by the position of the head). You may consider a 'push' or a 'pull' 
mechanism at this stage 

6 Consider whether any atom that has been changed now has too many bonds; if so one of them 
must be broken to avoid a ridiculous structure. Select a bond to break. Draw a curly arrow from 
the centre of the chosen bond, the filled orbital, and terminate it in a suitable place 

7 Write out the structures of the products specified by the curly arrows. Break the bonds that are 
the sources of the arrows and make those that are the targets. Consider the effect on the charges 
on individual atoms and check that the overall charge is not changed. Once you have drawn the 
curly arrows, the structure of the products is already decided and there is no room for any further 
decisions. Just write what the curly arrows tell you. If the structure is wrong, then the curly 
arrows were wrong so go back and change them 

8 Repeat stages 5-7 as required to produce a stable product 



Problems 



133 



When you have read through all the different types of reaction mechanism, practise drawing them 
out with and without the help of the book. Complete the exercises at the end of the chapter and then 
try to devise mechanisms for other reactions that you may know. You now have the tools to draw out 
in the universal pictorial language of organic chemists virtually all the mechanisms for the reactions 
you will meet in this book and more besides! 

Problems 

1. Each of these molecules is electrophilic. Identify the elec- 
trophilic atom and draw a mechanism for reaction with a general- 
ized nucleophile Nu~, giving the product in each case. 



3. Complete these mechanisms by drawing the structure of the 
products in each case. 



r 





°^° 



CI CI 



Me^ ^Cl 



MeOr ©N)Me 



2. Each of these molecules is nucleophilic. Identify the elec- 
trophilic atom and draw a mechanism for reaction with a general- 
ized electrophile E + , giving the product in each case. 



© 



H H 

MeO" 0^OMe 



H 2 N — NH 2 



h/> h -S^c1 



(b) 



Br 



> CG 



4. Each of these electrophiles could react with a nucleophile at (at 
least) two different atoms. Identify these atoms and draw a mech- 
anism for each reaction together with the products from each. 



I 



u Ph Ph 

H y 

Prr^© 



OH 



n 



© 



OMe 

I 

MeO^ N)Me 



NH 

X 



5. Put in the arrows on these structures (which have been drawn with all the atoms in the right places!) to give the products shown. 




(b) 



6. Draw mechanisms for these reactions. The starting materials 
have not necessarily been drawn in a helpful way. 

NaOH 



H 2 



EtCH 2 SH 



7. Draw a mechanism for this reaction. 
PhCHBr.CHBr.C0 2 H + NaHC0 3 



PhCH=CHBr 



NaOH, H 2 



Hints. First draw good diagrams of the reagents. NaHCC>3 is a salt 
and a weak base — strong enough only to remove which proton? 
Then work out which bonds are formed and which broken, decide 
whether to push or pull, and draw the arrows. What are the other 
products? 



(c) 




"CH, 



HBr 




OH 



Nucleophilic addition to the 
carbonyl group 



6 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• Functional groups, including the 


• How and why the C=0 group reacts 


• Additions of organometallic reagents 


carbonyl group (C=0) ch2 


with nucleophiles 


ch9 


• Identifying the functional groups in a 


• Explaining the reactivity of the C=0 


• C=0 groups with an adjacent double 


molecule spectroscopically ch3 


group using molecular orbitals and 


bond chlO 


• How molecular orbitals explain 


curly arrows 


• How the C=0 group in derivatives of 


molecular shapes and functional 


• What sorts of molecules can be made 


carboxylic acids promotes 


groups ch4 


by reactions of C=0 groups 


substitution reactions chl2 


• How, and why, molecules react 


• How acid or base catalysts improve 


• Substitution reactions of the C=0 


together, the involvement of functional 


the reactivity of the C=0 group 


group's oxygen atom chl4 


groups, and using curly arrows to 






describe reactions ch5 







Molecular orbitals explain the reactivity of the carbonyl group 

We are now going to leave to one side most of the reactions you met in the last chapter — we will 
come back to them all again later in the book. In this chapter we are going to concentrate on just one 
of them — probably the simplest of all organic reactions — the addition of a nucleophile to a carbonyl 
group. The carbonyl group, as found in aldehydes, ketones, and many other compounds, is without 
doubt the most important functional group in organic chemistry, and that is another reason why we 
have chosen it as our first topic for more detailed study. 

You met nucleophilic addition to a carbonyl group on p. 114 and 119, where we showed you how 
cyanide reacts with acetone to give an alcohol. As a reminder, here is the reaction again, with its 
mechanism. 

NaCN, H 2 S0 4 \ 
»► NC -A OH 



H 2 



alcohol 

("acetone cyanohydrin") 

78% yield 



NC 



nya. 



NC- 



nucleophilic addition of CN to 
the carbonyl group 



^o<O h © 



protonation 



The reaction has two steps: nucleophilic addition of cyanide, followed by protonation of the 
anion. In fact, this is a general feature of all nucleophilic additions to carbonyl groups. 

Additions to carbonyl groups generally consist of two mechanistic steps: 

1 Nucleophilic attack on the carbonyl group 

2 Protonation of the anion that results 



The addition step is more important, and it forms a new C-C bond at the expense of the C=0 7t 
bond. The protonation step makes the overall reaction addition of HCN across the C=0 Jt bond. 



We will frequently use a device 
like this, showing a reaction 
scheme with a mechanism for the 
same reaction looping round 
underneath. The reagents and 
conditions next to the arrow 
across the top will tell you how 
you might carry out the reaction, 
and the pathway shown 
underneath will tell you how it 
actually works. 



136 



6 ■ Nucleophilic addition to the carbonyl group 



Why does cyanide, in common with many other nucleophiles, attack the carbonyl group? And why 
does it attack the carbon atom of the carbonyl group? To answer these questions we need to look in detail 
at the structure of carbonyl compounds in general and the orbitals of the C=0 group in particular. 

The carbonyl double bond, like that found in alkenes (whose bonding we discussed in Chapter 4), 
consists of two parts: one O bond and one 7t bond. The bond between the two sp hybridized 
atoms — carbon and oxygen — is formed from two sp orbitals. The other sp orbitals on carbon form 
the two bonds to the substituents while those on oxygen are filled by the two lone pairs. The sp 
hybridization means that the carbonyl group has to be planar, and the angle between the substituents 
is close to 120°. The diagram illustrates all this for the simplest carbonyl compound, formaldehyde 
(or methanal, CH2O). The Jt bond then results from overlap of the remaining p orbitals — again, you 
can see this for formaldehyde in the diagram. 



n 



formaldehyde 
(methanal, CH 2 0) 



You were introduced to the polarization 
of orbitals in Chapter 4, and we 
discussed the case of the carbonyl 
group on p. 103. 



a bonding in formaldehyde 



remaining sp 
orbitals on C 120 
form bonds to H 



C-0 o bond made up 
of two sp 2 orbitals 




remaining sp 

orbitals on 

contain lone pairs 



% bonding in formaldehyde 



C-0 7t bond 

made up of two 

p orbitals 



Notice that we have drawn the Jt bond skewed towards oxygen. This is because oxygen is more 
electronegative than carbon, polarizing the orbital as shown. Conversely, the unfilled jc* antibond- 
ing orbital is skewed in the opposite direction, with a larger coefficient at the carbon atom. Put all of 
this together and we get the complete picture of the orbitals of a carbonyl group. 




empty, antibonding it* orbital 



R 

filled o sp 2 (lone pair) orbitals 





filled 71 orbital 




complete diagram of filled 
orbitals of C=0 bond 



Electronegativities, bond lengths, and bond strengths 



Representative bond energies, kJ mol 
C-0 351 C=0 720 



Representative bond lengths, A 
C-0 1.43 C=0 1.21 



Electronegativity 

C 2.5 



3.5 



Because there are two types of bonding between C and O, the C=0 double bond is rather shorter 
than a typical C-O single bond, and also over twice as strong — so why is it so reactive? Polarization is 
the key. The polarized C=0 bond gives the carbon atom some degree of positive charge, and this 
charge attracts negatively charged nucleophiles (like cyanide) and encourages reaction. The polariza- 
tion of the antibonding %* orbital towards carbon is also important, because, when the carbonyl 
group reacts with a nucleophile, electrons move from the HOMO of the nucleophile (an sp orbital in 
this case) into the LUMO of the electrophile — in other words the 71* orbital of the C=0 bond. The 
greater coefficient of the %* orbital at carbon means a better HOMO-LUMO interaction, so this is 
where the nucleophile attacks. 

As our nucleophile — which we are representing here as 'Nu~' — approaches the carbon atom, the 
electron pair in its HOMO starts to interact with the LUMO (antibonding JC*) to form a new bond. 



Cyanohydrins from the attack of cyanide on aldehydes and ketones 



137 



Filling antibonding orbitals breaks bonds and, as the electrons enter the antibonding 71* of the car- 
bonyl group, the 71 bond is broken, leaving only the C-O bond intact. But electrons can't just vanish, 
and those that were in the 7t bond move off on to the electronegative oxygen, which ends up with the 
negative charge that started on the nucleophile. You can see all this happening in the diagram below. 

curly arrow representation: 

Nu<T 



Nil 



^ 




,© 



The HOMO of the nucleophile will 
depend on what the nucleophile 
is, and we will meet examples in 
which it is an sp or sp 3 orbital 
containing a lone pair, or a B-H or 
metal-carbon a orbital. We shall 
shortly discuss cyanide as the 
nucleophile; cyanide's HOMO is 
an sp orbital on carbon. 



orbitals involved: 




n 



sp hybridized carbon 



Nu 



Nu 



HOMO 

e 




electrons in HOMO begin to 
interact with LUMO 



^ new a bond 



sp hybridized carbon 



while at the same time. 




filling of Jt* causes 
it bond to break 



u 



electrons from it bond end up 
as negative charge on oxygen 



Notice how the trigonal, planar sp hybridized carbon atom of the carbonyl group changes to a 
tetrahedral, sp hybridized state in the product. For each class of nucleophile you meet in this chap- 
ter, we will show you the HOMO-LUMO interaction involved in the addition reaction. 



Cyanohydrins from the attack of cyanide on aldehydes and 
ketones 

Now that we've looked at the theory of how a nucleophile attacks a carbonyl group, let's go back to 
the real reaction with which we started this chapter: cyanohydrin formation from a carbonyl com- 
pound and sodium cyanide. Cyanide contains sp hybridized C and N atoms, and its HOMO is an sp 
orbital on carbon. The reaction is a typical nucleophilic addition reaction to a carbonyl group: the 
electron pair from the HOMO of the CN~ (an sp orbital on carbon) moves into the C=0 Jt* orbital; 
the electrons from the C=0 71 orbital move on to the oxygen atom. The reaction is usually carried out 
in the presence of acid, which protonates the resulting alkoxide to give the hydroxyl group of the 
composite functional group known as a cyanohydrin. The reaction works with both ketones and 
aldehydes, and the mechanism below shows the reaction of a general aldehyde. 
NaCN 



CN 



R^H 

aldehyde 



OE 



H 2 0, HCI 



NC OH 

R^H 

cyanohydrin 

k 



^ 



C© HOMO = sp orbital 




"X 



0^ 

f J§> 



LUMO = Jt* 

orbitals involved in the additior 
of cyanide 



orbitals of the cyanide ion 
• N^=c e 



C-N a orbital 
(not shown) 



sp orbital on 
N contains 
lone pair 



HOMO = sp 
orbital on C 
containing 
lone pair 



w 



two pairs of p orbitals make 
two orthogonal jtbonds 



138 



6 ■ Nucleophilic addition to the carbonyl group 



Cyanohydrins in synthesis 

Cyanohydrins are important synthetic 
intermediates — for example, the cyanohydrin 
formed from this cyclic amino ketone forms the first 
step of a synthesis of some medicinal compounds 
known as 5HT3 agonists, which were designed to 
reduce nausea in chemotherapy patients. 
Cyanohydrins are also components of many natural 
and industrial products, such as the insecticides 
cypermethrin (marketed as 'Ripcord', 'Barricade', 
and 'Imperator') and fluvalinate. 




OH 



95% yield 



-CN 




OPh cypermethrin 



This is because the cyanide is a 
good leavinggroup — we'll come 
back to this type of reaction in 
much more detail in Chapter 12. 



Cyanohydrin formation is reversible: just dissolving a cyanohydrin in water can give back the alde- 
hyde or ketone you started with, and aqueous base usually decomposes cyanohydrins completely. 

HO CN NaOH, H 2 



R "R 

cyanohydrin 



OH 



°1 



R 



CN 



R "R 

ketone 



Q 0^ CN 
R^R 



CN 








sp 



NaCN 



^V H 



R H 2 0, HCI n ^_X 



120° 



R R 

109° 

substituents move 
closer together 



Cyanohydrin formation is therefore 
an equilibrium between starting materi- 
als and products, and we can only get 
good yields if the equilibrium favours 
the products. The equilibrium is more 
favourable for aldehyde cyanohydrins 
than for ketone cyanohydrins, and the 
reason is the size of the groups attached 



Some equilibrium constants 





JL 



HCN 



K e , 



NC OH 



aldehyde or 
ketone K e[ 



PhCHO 




28 



Cyanohydrins and cassava 

The reversibility of cyanohydrin formation is of more than 
theoretical interest. In parts of Africa the staple food is 
cassava. This food contains substantial quantities of the 
glucoside of acetone cyanohydrin (a glucoside is an acetal 
derived from glucose). We shall discuss the structure of 
glucose later in this chapter, but for now, just accept that 
it stabilizes the cyanohydrin. 

The glucoside is not poisonous in itself, but enzymes in 
the human gut break it down and release HCN. Eventually 
50 mg HCN per 100 g of cassava can be released and this 



is enough to kill a human being after a meal of 
unfermented cassava. If the cassava is crushed with 
water and allowed to stand ('ferment'), enzymes in the 
cassava will do the same job and then the HCN can be 
washed out before the cassava is cooked and eaten. 

The cassava is now safe to eat but it still contains some 
glucoside. Some diseases found in eastern Nigeria can be 
traced to long-term consumption of HCN. Similar 
glucosides are found in apple pips and the kernels inside 
the stones of fruit such as peaches and apricots. Some 
people like eating these, but it is unwise to eat too many 
atone sitting! 




glucoside of acetone 
cyanohydrin found in 
cassava 



OH 



A 



(i-glucosidase 



hydroxynitrile 
lyase 

■*■ HC^JJN *- 

(an enzyme) /\ (another enzyme) 



+ HCN 



Nucleophilic attack by 'hydride' on aldehydes and ketones 



139 



to the carbonyl carbon atom. As the carbonyl carbon atom changes from sp to sp , its bond angles 
change from about 120° to about 109° — in other words, the substituents it carries move closer 
together. This reduction in bond angle is not a problem for aldehydes, because one of the sub- 
stituents is just a (very small) hydrogen atom, but for ketones, especially ones that carry larger alkyl 
groups, this effect can disfavour the addition reaction. Effects that result from the size of substituents 
and the repulsion between them are called steric effects, and we call the repulsive force experienced 
by large substituents steric hindrance. 



Steric hindrance (not hinderance) 
is a consequence of repulsion 
between the electrons in all the 
filled orbitals of the alkyl 
substituents. 



The angle of nucleophilic attack on aldehydes and ketones 

Having introduced you to the sequence of events that makes up a nucleophilic attack at C=0 (inter- 
action of HOMO with LUMO, formation of new bond, breakage of % bond), we should now tell 
you a little more about the direction from which the nucleophile approaches the carbonyl group. Not 
only do nucleophiles always attack carbonyl groups at carbon, but they also always approach from a 
particular angle. You may at first be surprised by this angle, since nucleophiles attack not from a 
direction perpendicular to the plane of the carbonyl group but at about 107° to the C=0 bond. This 
approach route is known as the Biirgi-Dunitz trajectory after the authors of the elegant crystallo- 
graphic methods that revealed it. You can think of the angle of attack as the result of a compromise 
between maximum orbital overlap of the HOMO with 71* and minimum repulsion of the HOMO by 
the electron density in the carbonyl 71 bond. 

N„© 



(J 7t* 



-N 



maximum overlap with it* 
perpendicular to C=0 bond 



repulsion from filled n orbital 
m„© forces nucleophile to attack at 

obtuse angle 



nucleophile attacks 
C=0 at 107° angle 



Nu 



> 



combined effect: 







107° 



^. 




u 



J 



Any other portions of the molecule that get in the way of (or, in other words, that cause steric hin- 
drance to) the Biirgi-Dunitz trajectory will greatly reduce the rate of addition and this is another rea- 
son why aldehydes are more reactive than ketones. The importance of the Biirgi-Dunitz trajectory 
will become more evident later — particularly in Chapter 34. 

Nucleophilic attack by 'hydride' on aldehydes and ketones 

Nucleophilic attack by the hydride ion, H~, is not a known reaction. This species, which is present in 
the salt sodium hydride, NaH, is so small and has such a high charge density that it only ever reacts 
as a base. The reason is that its filled Is orbital is of an ideal size to interact with the hydrogen 






H 







Mer s Me 



H^ 



H-7-X 



-*- H 2 + X 







V 



nucleophilic attack by H 
never happens 



H always reacts as a base 



■The Burgi-Dunitz angle 
Burgi and Dunitz deduced this 
trajectory by examining crystal 
structures of compounds containing 
both a nucleophilic nitrogen atom and 
an electrophilic carbonyl group. They 
found that, when the two got close 
enough to interact, but were not free to 
undergo reaction, the nitrogen atom 
always lay on or near the 107° 
trajectory decribed here. Theoretical 
calculations later gave the same 107° 
value for the optimum angle of attack. 



Although we now know precisely 
from which direction the 
nucleophile attacks the C=0 
group, this is not always easy to 
represent when we draw curly 
arrows. As long as you bear the 
Burgi-Dunitz trajectory in mind, 
you are quite at liberty to write any 
of the variants shown here , 
among others. 



■.<n|£ 



J? 



Nu©^\ 



140 



6 ■ Nucleophilic addition to the carbonyl group 



atom's contribution to the o"* orbital of an H-X bond (X can be any atom), but much too small to 
interact easily with carbon's more diffuse 2p orbital contribution to the LUMO (Jt*) of the C=0 
group. 

Nevertheless, adding H to the carbon atom of a C=0 group would be a very useful reaction, as 
the result would be the formation of an alcohol. This process would involve going down from the 
aldehyde or ketone oxidation level to the alcohol oxidation level (Chapter 2, pp. 25-36) and would 
therefore be a reduction. It cannot be done with NaH, but it can be done with some other com- 
pounds containing nucleophilic hydrogen atoms. 

reduction of a ketone to an alcohol 



Mer^Me 



H OH 



HO® „ H 



Me Me 



In Chapter 4, we looked at isoelectronic 
BH3 and CH3. Here, we have effectively 
just added H~to both of them. 



H H 

borohydride anion 



H H 

methane 



Just as we have used Nu~ to indicate 
any (undefined) nucleophile, here E + 
means any (undefined) electrophile. 



The reason that H~ never acts as 
a nucleophile is that its Is orbital 
is too small. The orbitals involved 
in borohydride reductions are the 
7t* of the C=0 group as the LUMO 
and a B-H O orbital as the HOMO, 
so there is a much better orbital 
match. 



H H 






HOMO = B-H a 



8= 



The most important of these compounds is sodium borohydride, NaBH^ This is a water-soluble 
salt containing the tetrahedral BH4 anion, which is isoelectronic with methane but has a negative 
charge since boron has one less proton in the nucleus than does carbon. 

But beware! The boron's negative charge doesn't mean that there is a lone pair on boron — there 
isn't. You cannot draw an arrow coming out of this charge to form another bond. If you did, you 
would get a pentacovalent B(V) compound, which would have 10 electrons in its outer shell. Such a 
thing is impossible with a first row element as there are only four available orbitals ( 1 x 2s and 3 x 
2p). Instead, since all of the electrons (including that represented by the negative charge) are in B-H 
orbitals, it is from a B-H bond that we must start any arrow to indicate reaction of BH4 as a nucle- 
ophile. By transferring this pair of electrons we make the boron atom neutral — it is now trivalent 
with just six electrons. 



arrow cannot start on negative 
charge: no lone pair on B 



electrons must be 
transferred from a bond 



)b<0 E@ 



eight electrons 
in B-H bonds 



*- 



H H 

\/ 

H 



H 




impossible structure: ten 
electrons in B-H bonds 



eight electrons in 
B-H bonds 



H 

six electrons in B-H 

bonds and one empty p 

orbital 



What happens when we carry out this reaction using a carbonyl compound as the electrophile? 
The hydrogen atom, together with the pair of electrons from the B-H bond, will be transferred to the 
carbon atom of the C=0 group. 




I 
H 



H 0© 

X. 



Though no hydride ion, H~, is actually involved in the reaction, the transfer of a hydrogen atom 
with an attached pair of electrons can be regarded as a 'hydride transfer'. You will often see it 
described this way in books. But be careful not to confuse BH4 with the hydride ion itself. To make it 
quite clear that it is the hydrogen atom that is forming the new bond to C, this reaction may also be 
helpfully represented with a curly arrow passing through the hydrogen atom. 




H 0© 

X. 



The oxyanion produced in the first step can help stabilize the electron-deficient BH3 molecule by 
adding to its empty p orbital. Now we have a tetravalent boron anion again, which could transfer a 
second hydrogen atom (with its pair of electrons) to another molecule of aldehyde. 



Nucleophilic attack by 'hydride' on aldehydes and ketones 



141 



■S,/» 



yf^l 



H H n - 



X 



pr^e<H 



H 



H 

\ 



H^R 



/ B - H u „© 



X X 



H 1 - 



This process can continue so that, in principle, all four hydrogen atoms could be transferred to 
molecules of aldehyde. In practice the reaction is rarely as efficient as that, but aldehydes and ketones 
are usually reduced in good yield to the corresponding alcohol by sodium borohydride in water or 
alcoholic solution. The water or alcohol solvent provides the proton needed to form the alcohol 
from the alkoxide. 



examples of reductions with sodium borohydride 



Me 



NaBH 4 



OH 



H 2 



Me 






NaBH 4 



/'-PrOH 




Sodium borohydride is one of the weakest hydride donors available. The fact that it can be used in 
water is evidence of this as more powerful hydride donors such as lithium aluminium hydride, 
L1AIH4, react violently with water. Sodium borohydride reacts with both aldehydes and ketones, 
though the reaction with ketones is slower: for example, benzaldehyde is reduced about 400 times 
faster than acetophenone in isopropanol. 

Sodium borohydride does not react at all with less reactive carbonyl compounds such as 
esters or amides: if a molecule contains both an aldehyde and an ester, only the aldehyde will be 
reduced. 




The next two examples illustrate the reduction of aldehydes and ketones in the presence of other 
reactive functional groups. No reaction occurs at the nitro group in the first case or at the alkyl halide 
in the second. 







PIT ^H 

benzaldehyde 



PIT Me 

acetophenone 



Aluminium is more electropositive 
(more metallic) than boron and is 
therefore more ready to give up a 
hydrogen atom (and the 
associated negative charge), 
whether to a carbonyl group or to 
water. Lithium aluminium hydride 
reacts violently and dangerously 
with water in an exothermic 
reaction that produces highly 
flammable hydrogen. 



© 



H H 



violent 
reaction! 



a 



OH 



\ 



142 



6 ■ Nucleophilic addition to the carbonyl group 



Addition of organometallic reagents to aldehydes and ketones 



Organometallic compounds 
always have a metal-carbon 
bond. 



Electronegativities 



C 
2.5 



Li 
1.0 



Mg 
1.2 



We explained on p. 102 the origin of 
the polarization of bonds to 
electropositive elements. 



■Jt. 



J© 

HOMO = Li-C a 
polarized towards C 




LUMO = 7t* 

orbitals involved in the addition 
of methyllithium 



organometallics are 
destroyed by water 

-Me ' H-LOH 



fast and 
exothermic 



Me H 

methane 



LiOH 



Aprotic solvents contain no acidic 
protons, unlike, say, water or 
alcohols. 



'Secondary' and 'tertiary' alcohols are 
defined on p. 30. 



The next type of nucleophile we shall consider is the organometallic reagent. Lithium and magne- 
sium are very electropositive metals, and the Li-C or Mg-C bonds in organolithium or organomag- 
nesium reagents are highly polarized towards carbon. They are therefore very powerful nucleophiles, 
and attack the carbonyl group to give alcohols, forming a new C-C bond. For our first example, we 
shall take one of the simplest of organolithiums, methyllithium, which is commercially available as a 
solution in Et20, shown here reacting with an aldehyde. The orbital diagram of the addition step 
shows how the polarization of the C-Li bond means that it is the carbon atom of the nucleophile that 
attacks the carbon atom of the electrophile and we get a new C-C bond. 



X 



OH 



1. MeLi, THF 

* 

2. H 2 



M 





0^ 



H-*-0H 



-Me 



Me 



The course of the reaction is much the same as you have seen before, but we need to highlight a 
few points where this reaction scheme differs from those you have met earlier in the chapter. First of 
all, notice the legend '1. MeLi, THF; 2. H2O'. This means that, first, MeLi is added to the aldehyde in 
a THF solvent. Reaction occurs: MeLi adds to the aldehyde to give an alkoxide. Then (and only then) 
water is added to protonate the alkoxide. The '2. H 2 0' means that water is added in a separate step 
only when all the MeLi has reacted: it is not present at the start of the reaction as it was in the cyanide 
reaction and some of the borohydride addition reactions. In fact, water must not be present during 
the addition of MeLi (or of any other organometallic reagent) to a carbonyl group because water 
destroys organometallics very rapidly by protonating them to give alkanes (organolithiums 
and organomagnesiums are strong bases as well as powerful nucleophiles). The addition of 
water, or sometimes dilute acid or ammonium chloride, at the end of the reaction is known as the 
work-up. 

Because they are so reactive, organolithiums are usually reacted at low temperature, often -78 °C 
(the sublimation temperature of solid CO2), in aprotic solvents such as Et20 or THF. Organo- 
lithiums also react with oxygen, so they have to be handled under a dry, inert atmosphere of nitrogen 
or argon. 

Other common, and commercially available, organolithium reagents include n-butyllithium and 
phenyllithium, and they react with both aldehydes and ketones. Note that addition to an aldehyde 
gives a secondary alcohol while addition to a ketone gives a tertiary alcohol. 





I 

R' "H 






1. PhLi, THF 

1 

2. H 2 



OH 



PrT / ^H 
R 

secondary alcohol 






1. n-BuLi, THF 

»■ 

2. H 2 



OH 




tertiary alcohol 



Victor Grignard (1871-1935) of the 
University of Lyon was awarded the 
Nobel Prize for chemistry in 1912 for 
his discovery of these reagents. 



Organomagnesium reagents known as Grignard reagents (RMgX) react in a similar way. Some 
simple Grignard reagents, such as methyl magnesium chloride, MeMgCl, and phenyl magnesium 
bromide, PhMgBr, are commercially available, and the scheme shows PhMgBr reacting with an alde- 
hyde. The reactions of these two classes of organometallic reagent — organolithiums and Grignard 
reagents — with carbonyl compounds are among the most important ways of making carbon-carbon 
bonds, and we will consider them in more detail in Chapter 9. 



Addition of water to aldehydes and ketones 



143 



O OH 

AX. PhMgBr, Et 2 
H *- PlW^H 



2. H 2 



BrMg-^Ph 



R "H 



r~^ hJ-oh 

0© 



PIT / H 
R 




Addition of water to aldehydes and ketones 

Nucleophiles don't have to be highly polarized or negatively charged to react with aldehydes and 
ketones: neutral ones will as well. How do we know? This C NMR spectrum was obtained by dis- 
solving formaldehyde, H2C=0, in water. You will remember from Chapter 3 that the carbon atoms 
of carbonyl groups give C signals typically in the region of 150-200 p.p.m. So where is formalde- 
hyde's carbonyl peak? Instead we have a signal at 83 p.p.m. — where we would expect tetrahedral car- 
bon atoms singly bonded to oxygen to appear. 



13 C NMR spectrum of formaldehyde in water 



200 



150 



100 



50 



What has happened is that water has added to the carbonyl group to give a compound known as a 



hydrate or 1,1-diol. 



expect C signal 
between 150 and 
200 p.p.m. 



JL 



HQ OH 



H 2 



X: 



13 C signal at 
83 p.p.m. 



formaldehyde hydrate or 1,1-diol 

This reaction, like the cyanohydrin formation we discussed at the beginning of the chapter, is an 
equilibrium, and is quite general for aldehydes and ketones. But, as with the cyanohydrins, the posi- 
tion of the equilibrium depends on the structure of the carbonyl compound. Generally, the same 
steric factors (pp. 138-139) mean that simple aldehydes are hydrated to some extent while simple 
ketones are not. However special factors can shift the equilibrium towards the hydrated form even 
for ketones, particularly if the carbonyl compound is reactive or unstable. 

Formaldehyde is an extremely reactive aldehyde as it has no substituents to hinder attack — it is so 
reactive that it is rather prone to polymerization (Chapter 52). And it is quite happy to move from 
sp to sp hybridization because there is very little increased steric hindrance between the two hydro- 
gen atoms as the bond angle changes from 120° to 109° (p. 139). This is why our aqueous solution of 
formaldehyde contains essentially no CH2O — it is completely hydrated. A mechanism for the hydra- 
tion reaction is shown below. Notice how a proton has to be transferred from one oxygen atom to 
the other, mediated by water molecules. 



.O 



h/\J£ 



"4? 



!X 



OH, 



0© 



HO.) 



^ H V C 



,0© 



D 



H'^^H 



^ H \/ 



OH 



H""^H 



u 



H 2 



HO OH 



significant concentrations of 

hydrate are generally formed 

only from aldehydes 



HOMO = oxygen sp J 




orbital containing 
one pair 



LUMO: 



orbitals involved in the 
addition of water 



144 



6 ■ Nucleophilic addition to the carbonyl group 



Monomeric formaldehyde 

The hydrated nature of formaldehyde poses a problem for chemistry that requires 
anhydrous conditions such as the organometallic additions we have just been talking 
about. Fortunately crac/</ng(heatingto decomposition) the polymeric 'paraformaldehyde' 
can provide monomeric formaldehyde in anhydrous solution. 



polymeric 'paraformaldehyde' 



OH 



HO" I ^0 
CH 2 



Chloral hydrate is the infamous 'knock 
out drops' of Agatha Christie or the 
'Mickey Finn' of Prohibition gangsters. 



Formaldehyde reacts with water so readily because its substituents are very small: a steric effect. 
Electronic effects can also favour reaction with nucleophiles — electronegative atoms such as halo- 
gens attached to the carbon atoms next to the carbonyl group can increase the extent of hydration 
according to the number of halogen substituents and their electron-withdrawing power. They 
increase the polarization of the carbonyl group, which already has a positively polarized carbonyl 
carbon, and make it even more prone to attack by water. Trichloroacetaldehyde (chloral, CI3CHO) 
is hydrated completely in water, and the product 'chloral hydrate' can be isolated as crystals and is an 
anaesthetic. You can see this quite clearly in the two IR spectra. The first one is a spectrum of chloral 
hydrate from a bottle — notice there is no strong absorption between 1700 and 1800 cm~ (where we 
would expect C=0 to appear) and instead we have the tell-tale broad O-H peak at 3400 crrT . 
Heating drives off the water, and the second IR spectrum is of the resulting dry chloral: the C=0 
peak has reappeared at 1770 cm~ , and the O-H peak has gone. 
IR spectrum of chloral (nujol) IR spectrum of chloral hydrate (nujol) 

' 100- 





4000 v 3000 



IR spectrum of chloral (nujol) 



4000 v 3000 2000 1500 

IR spectrum of chloral hydrate (nujol) 



1000 



400 



The chart shows the extent of hydration (in water) of a small selection of carbonyl compounds: 
hexafiuoroacetone is probably the most hydrated carbonyl compound possible! 





J 



X + H 2 



HO OH 
R^R 



acetone 



acetaldehyde 



formaldehyde 



X. 



equilibrium constant K 
0.001 

1.06 



2280 



chloral y 


hci 

CI 



2000 


hexafiuoroacetone ^x^ 


-V F 


1 200 000 


f^t 


>F 





Cyclopropanones — three-membered ring ketones — are also hydrated to a significant extent, but 
for a different reason. You saw earlier how acyclic ketones suffer increased steric hindrance when the 
bond angle changes from 120° to 109° on moving from sp to sp hybridization. Cyclopropanones 



Hemiacetals from reaction of alcohols with aldehydes and ketones 



145 



(and other small-ring ketones) conversely prefer the small bond angle because their substituents are 
already confined within a ring. Look at it this way: a three-membered ring is really very strained, with 
bond angles forced to be 60°. For the sp hybridized ketone this means bending the bonds 60° away 
from their 'natural' 120°. But for the sp hybridized hydrate the bonds have to be distorted by only 
49° (= 109° - 60°). So addition to the C=0 group allows some of the strain inherent in the small ring 
to be released — hydration is favoured, and indeed cyclopropanone and cyclobutanone are very reac- 
tive electrophiles. 

The same structural features that favour or disfavour hydrate formation are 
important in determining the reactivity of carbonyl compounds with other 
nucleophiles, whether the reactions are reversible or not. Steric hindrance and 
more alkyl substituents make carbonyl compounds less reactive towards any 
nucleophile; electron-withdrawing groups and small rings make them more 
reactive. 



cyclopropanone 
O 
sp 2 C wants 120°, I 
but gets 60° A. 



H 2 



HO OH 

sp 3 C wants 109°, \V 
but gets 60° l\ 

cyclopropanone hydrate 



Hemiacetals from reaction of alcohols with aldehydes and 
ketones 

Since water adds to (at least some) carbonyl compounds, it should come as no surprise that alcohols 
do too. The product of the reaction is known as a hemiacetal, because it is halfway to an acetal, a 
functional group, which you met in Chapter 2 (p. 35) and which will be discussed in detail in 
Chapter 14. The mechanism follows in the footsteps of hydrate formation: just use ROH instead of 
HOH. 



R^H 
aldehyde 



Et^OH | \-^ 
R^H 



EtOH 



EtQ OH 

R^H 

hemiacetal 



( H 

r 



HOEt 



Et ^ 7 ,oe 



- e :v 



.00 



OEt 



A proton has to be transferred from one oxygen atom to the other: we have shown ethanol doing 
this job, with one molecule being protonated and one deprotonated. There is no overall consump- 
tion of ethanol in the protonation/deprotonation steps, and the order in which these steps happen is 
not important. In fact, you could reasonably write them in one step as shown in the margin, without 
involving the alcohol, and we do this in the next hemiacetal-forming reaction below. As with all these 
carbonyl group reactions, what is really important is the addition step, not what happens to the pro- 
tons. 



Hemiacetal formation is reversible, and 
hemiacetals are stabilized by the same special 
structural features as those of hydrates. 
However, hemiacetals can also gain stability 
by being cyclic — when the carbonyl group 
and the attacking hydroxyl group are part of 
the same molecule. The reaction is now an 
intramolecular (within the same molecule) 
addition, as opposed to the intermolecular 
(between two molecules) ones we have con- 
sidered so far. 



- rV 0H 



— ^V^H 



hydroxyaldehyde 



cyclic hemiacetal 




intramolecular attack of 
hydroxyl group 






R 2 0. OH 



R 2 0. OR 3 



^\u D l/\ 



"H 

hemiacetal 

R 2 OH 



acetal 



OH 




H 



hemiacetal from a cyclic 

ketone hemiacetal 

(or "hemiketal") (or "lactol") 

names for functional groups 



H>0 

51? ■ 



EtQ OH 



R^^^H 



■ /ntemiolecular reactions occur 
between two molecules 

■ /ntramolecular reactions occur 
within the same molecule 



We shall discuss the reasons why 
intramolecular reactions are more 
favourable, and why cyclic hemiacetals 
and acetals are more stable, in Chapter 
14. 



146 



6 ■ Nucleophilic addition to the carbonyl group 



Although the cyclic hemiacetal (also called 'lactol') product is more stable, it is still in equi- 
librium with some of the open-chain hydroxyaldehyde form. Its stability, and how easily it 
forms, depend on the size of the ring: five- and six-membered rings are free from strain (their bonds 
are free to adopt 109° or 120° angles — compare the three-membered rings on p. 145), and five- 
or six-membered hemiacetals are common. Among the most important examples are many 
sugars. Glucose, for example, is a hydroxyaldehyde that exists mainly as a six-membered cyclic hemi- 
acetal (>99% of glucose is cyclic in solution), while ribose exists as a five-membered cyclic 
hemiacetal. 



The way we have represented 
some of these molecules may be 
unfamiliartoyou: we have shown 
stereochemistry (whether bonds 
come out of the paper or into it — 
the wiggly lines indicate a mixture 
of both) and, forthe cyclic 
glucose, conformation (the actual 
shape the molecules adopt). 
These are very important in the 
sugars: we devote Chapter 16 to 
stereochemistry and Chapter 18 
to conformation. 




OH OH OH 

hydroxyaldehyde 



OH 



OH 



OH OH 

hydroxyaldehyde 



can be 
drawn as 



OH 



can be 
drawn as 




OH 

hydroxyaldehyde 

OH 




HO" 'OH 

hydroxyaldehyde 



HO v 



cyclic glucose: >99% 
in this form 



Vy 0H 

•> -5- 

^ '', 

HO OH 

cyclic ribose 



Ketones can form hemiacetals 



Hydroxyketones also form hemiacetals, but (as 
you should now expect) they usually do so less 
readily than hydroxyaldehydes. However, this 
hydroxyketone must exist solely as the cyclic 
hemiacetal because it shows no C=0 stretch in 
its IR spectrum. The reason? The starting 
hydroxyketone is already cyclic, with the hydroxyl 
group poised to attack the ketone — it can't get 
away, so cyclization is highly favoured. 





A catalyst increases the rate of a 
chemical reaction but emerges 
from the reaction unchanged. 



Acid and base catalysis of hemiacetal and hydrate formation 

In Chapter 8 we shall look in detail at acids and bases, but at this point we need to tell you about one 
of their important roles in chemistry: they act as catalysts for a number of carbonyl addition reac- 
tions, among them hemiacetal and hydrate formation. To see why, we need to look back at the mech- 
anisms of hemiacetal formation on p. 145 and hydrate formation on p. 143. Both involve 
proton-transfer steps like this, 
ethanol acting as a base 



Et- 



( HOEt 

^ft. ,0© 

R^H 



R^H 



oe^O 



OEt 



ethanol acting as an acid 



We introduced protonation by acid in 
Chapter 5, pp. 119-121. 



In the first proton-transfer step, ethanol acts as a base, removing a proton; in the second it acts as 
an acid, donating a proton. Strong acids or strong bases (for example, HC1 or NaOH) increase the 
rate of hemiacetal or hydrate formation because they allow these proton-transfer steps to occur 
before the addition to the carbonyl group. 



Acid and base catalysis of hemiacetal and hydrate formation 



147 



In acid (dilute HC1, say), this is the mechanism. The first step is now protonation of the carbonyl 
group's lone pair: the positive charge makes it much more electrophilic so the addition reaction is 
faster. Notice how the proton added at the beginning is lost again at the end — it really is a catalyst. In 
acid it is also possible for the hemiacetal to react further with the alcohol to form an acetal, but this 
need not concern you at present. 



hemiacetal formation in acid 



EtOH 



A. 



,-<V 



A. 



acid catalyst 



®y 



_^ Et— OH ] M 



FT^H 



EtQ OH 



(reactions discussed in 
Chapter 14) 



proton regenerated 



Et- 



^A. _,OH 
R^H 



EtQ OEt 



in acid solution, 
further reactions 
may take place 
leading to an 
acetal 



protonation makes carbonyl 
group more electrophilic 



And this is the mechanism in basic solution. The first step is now deprotonation of the ethanol by 
hydroxide, which makes the addition reaction faster by making the ethanol more nucleophilic. 
Again, base (hydroxide) is regenerated in last step, making the overall reaction catalytic in base. The 
reaction in base always stops with the hemiacetal — acetals never form in base. 



hemiacetal formation in base 



EtOH 






base catalyst 



EtQ OH 
R^H 



X -a- X 



EtQ OEt 
R-^-H 



acetals are never 
formed in base 



roH o 

J> * X 











base regenerated 


^ Et— 0© \ 


^H 


Et-. 




^OH 


R' 




deprotonation makes ethanol 
nucleophilic (as ethoxide) 


more 







The final step could equally well involve deprotonation of ethanol to give alkoxide — and 
alkoxide could equally well do the job of catalysing the reaction. In fact, you will often come 
across mechanisms with the base represented just as 'B - ' because it doesn't matter what the base 
is. 

These two mechanisms typify acid- and base-catalysed additions to carbonyl groups and we can 
summarize the effects of the two catalysts. 

# For nucleophilic additions to carbonyl groups: 

• Acid catalysts work by making the carbonyl group more electrophilic 

• Base catalysts work by making the nucleophile more nucleophilic 



EtO^ 



c% 



■*- EI^O 







148 



6 ■ Nucleophilic addition to the carbonyl group 



©O OH HOMO = sulfur 
n )c/ hybrid 



^ 



orbital containing 
lone pair 



/= 



orbitals involved in the 
addition of bisulfite 



The structure of NaHS0 3 , sodium 
bisulfite, is rather curious. It is an 
oxyanion of a sulfur(IV) compound 
with a lone pair of electrons — the 
HOMO — on the sulfur atom, but 
the charge is formally on the more 
electronegative oxygen As a 'third 
row' element (third row of the 
periodic table, that is) sulfur can 
have more that just eight 
electrons — it's all right to have 
four or six bonds to S or P, unlike, 
say, Bor N. 



Bisulfite addition compounds 



The last nucleophile of this chapter, sodium bisulfite, NaHS03, adds to aldehydes and some ketones 
to give what is usually known as a bisulfite addition compound. The reaction occurs by nucleophilic 
attack of a lone pair on the carbonyl group, just like the attack of cyanide. This leaves a positively 
charged sulfur atom but a simple proton transfer leads to the product. 



LUMO = 71* 



sodium bisulfite 
©, 



A / 0H 



// 








M 



<r^iv 



Me 



Me'^^Me 



Na®\/ 

^ ' 0H 



Me^^Me 



bisulfite addition compound 



The products are useful for two reasons. They are usually crystalline and so can be used to purify 
liquid aldehydes by recrystallization. This is of value only because this reaction, like several you have 
met in this chapter, is reversible. The bisulfite compounds are made by mixing the aldehyde or 
ketone with saturated aqueous sodium bisulfite in an ice bath, shaking, and crystallizing. After 
purification the bisulfite addition compound can be hydrolysed back to the aldehyde in dilute aque- 
ous acid or base. 



X 



stir together 
in ice bath 



NaHS0 3 ^_ 



dilute acid 
or base 



:y 



30 3 e Na® 



R "H 

crystalline solid 



The reversibility of the reaction makes bisulfite compounds useful intermediates in the synthesis 
of other adducts from aldehydes and ketones. For example, one practical method for making 
cyanohydrins involves bisulfite compounds. The famous practical book 'Vogel' suggests reacting 
acetone first with sodium bisulfite and then with sodium cyanide to give a good yield (70%) of the 
cyanohydrin. 

1. NaHS0 3 
2. NaCN HO CN 

*~ X 70% yield 

Me Me'^^Me 



What is happening here? The bisulfite compound forms first, but only as an intermediate on the 
route to the cyanohydrin. When the cyanide is added, reversing the formation of the bisulfite com- 
pound provides the single proton necessary to to give back the hydroxyl group at the end of the reac- 
tion. No dangerous HCN is released (always a hazard when cyanide ions and acid are present 
together). 



M«T^Me 



©, 



1. NaHS0 3 Na © \f 



Me^^Me 



J P H 
M«r^Me 



HO. CN 



2. NaCN 



M 



Me 



e 



„ o„ 



M 




Na 2 S0 3 
Q^ Q/ ^b CN 

r®cN — ^ v 

Me Me Me 



Bisulfite addition compounds 



149 



Other compounds from cyanohydrins 

Cyanohydrins can be converted by simple reactions 
into hydroxyacids or amino alcohols. Here is one 
example of each, but you will have to wait until 
Chapter 12 forthe details and the mechanisms of 
the reactions. Note that one cyanohydrin was made 
by the simplest method — just NaCN and acid — while 
the other came from the bisulfite route we have just 
discussed. 

amino alcohols by reduction of CN in cyanohydrin 



hydroxyacids by hydrolysis of CN in cyanohydrin 



y 



NaCN 



HO CN 



Me HCI, Et 2 Ph 



f~IVIe 



HCI 



H,0 



HO C0 2 H 
Pir^Me 







The bisulfite compound of formaldehyde (CH 2 0) has special significance. Earlier in this chapter 
we mentioned the difficulty of working with formaldehyde because it is either an aqueous solution 
or a dry polymer. One readily available monomeric form is the bisulfite compound. It can be made 
in water (in which it is soluble) but addition of ethanol (in which it isn't) causes it to crystallize 
out. 

aturated aqueous The compound is commercially available 

3 _/~\ © © an< ^' together with the related zinc salt, is 

j oe batn widely used in the textile industry as a reduc- 

crystalline ing agent 

The second reason that bisulfite compounds are useful is that they are soluble in water. Some 
small (that is, low molecular weight) aldehydes and ketones are water-soluble — acetone is an exam- 
ple. But most larger (more than four or so carbon atoms) aldehydes and ketones are not. This does 
not usually matter to most chemists as we often want to carry out reactions in organic solvents rather 
than water. But it can matter to medicinal chemists, who make compounds that need to be com- 
patible with biological systems. And in one case, the solubility of bisulfite adduct in water is literally 
vital. 

Dapsone is an antileprosy drug. It is a very effective one too, especially when used in combination 
with two other drugs in a 'cocktail' that can be simply drunk as an aqueous solution by patients in 
tropical countries without any special facilities, even in the open air. But there is a problem! Dapsone 
is insoluble in water. 

The solution is to make a bisulfite compound from it. You may ask how this is possible since dap- 
sone has no aldehyde or ketone — just two amino groups and a sulfone. The trick is to use the 
formaldehyde bisulfite compound and exchange the OH group for one of the amino groups in dap- 
sone. 






Y 




formaldehyde bisulfite 
adduct 

HO^^SO, Na© 



H 2 N ^^ ^^ ^NH 2 

dapsone: antileprosy drug; insoluble in water 



H 2 N 




water-soluble "pro-drug" 



Now the compound will dissolve in water and release dapsone inside the patient. The details of 
this sort of chemistry will come in Chapter 14 when you will meet imines as intermediates. But at this 
stage we just want you to appreciate that even the relatively simple chemistry in this chapter is useful 
in synthesis, in commerce, and in medicine. 



150 



6 ■ Nucleophilic addition to the carbonyl group 



Problems 



1. Draw mechanisms for these reactions. 
NaBH 4 



0° 

J> CHO 



EtOH, H 2 
LiAIH 4 



OH 



-cv 



OH 



2. Cyclopropanone exists as the hydrate in water but 
2-hydroxyethanal does not exist as its hemiacetal. Explain. 
HO OH 



1 -^ V 



cyclo- 
propanone 



hydrate 



HO 



^- 



OH 



3. One way to make cyanohydrins is illustrated here. Suggest a 
detailed mechanism for the process. 

H Me 3 Si — CN H CN 

„^n " D ^\r 



cat KCN 



^0SiMe 3 



4. There are three possible products from the reduction of this 
compound with sodium borohydride. What are their structures? 
How would you distinguish them spectroscopically, assuming 
you can isolate pure compounds? 




5. The triketone shown here is called 'ninhydrin' and is used for 
the detection of amino acids. It exists in aqueous solution as a 
monohydrate. Which of the three ketones is hydrated and why? 




6. This hydroxyketone shows no peaks in its infrared spectrum 
between 1600 and 1800 cm~ but it does show a broad absorption at 
3000 to 3400 cm -1 . In the 13 CNMR spectrum, v , O 
there are no peaks above 150 p.p.m. but there is HO v 
a peak at 110 p.p.m. Suggest an explanation. 

7. Each of these compounds is a hemiacetal and therefore formed 
from an alcohol and a carbonyl compound. In each case give the 
structure of these original materials. 

OH HO 




OH 




Me 




Me 



OH 




8. Trichloroethanol may be prepared by the direct reduction of 
chloral hydrate in water with sodium borohydride. Suggest a 
mechanism for this reaction. (Warning! Sodium borohydride 
does not displace hydroxide from carbon atoms!) 
HO OH N aBH 4 

y^ — *- ^ 

CI 3 C^ ^H 

chloral hydrate 



H,0 



CI 3 C^ ^OH 

trichloroethanol 



9. It has not been possible to prepare the adducts from simple 
aldehydes and HC1. What would be the structure of such com- 
pounds, if they could be made, and what would be the mechanism 
of their formation? Why cannot these compounds in fact be 
made? 

10. What would be the products of these reactions? In each case 
give a mechanism to justify your predictions. 

NaCN 
► ? 



v CHO 



H 2 0, HCI 





NaBH 4 



EtMgBr 

»~ ? 

Et 2 

11. The equilibrium constant K e „ for formation of the cyanohy- 
drin of cyclopentanone and HCN is 67, while for butan-2-one and 
HCN it is 28. Explain. 



Derealization and conjugation 




Connections 


Building on: 




Arriving at: 




Looking forward to: 


• Orbitals and bonding ch4 


• 


Interaction between orbitals over 


• 


Acidity and basicity chS 


• Representing mechanisms by curly 




many bonds 


• 


Conjugate addition and substitution 


arrows ch5 


• 


Stabilization by the sharing of 




chlO 


• Ascertaining molecular structure 




electrons over molecules 


• 


Chemistry of aromatic compounds 


spectroscopically ch3 


• 


Where colour comes from 




ch22 & ch23 




• 


Molecular shape and structure 


• 


Enols and enolates ch21, ch25-ch29 






determine reactivity 


• 


Chemistry of heterocycles ch43 & 




• 


Representing one aspect of structure 




ch44 






by curly arrows 


• 


Chemistry of life ch49-ch51 




• 


Structure of aromatic compounds 







Introduction 

As you look around you, you will be aware of many different colours — from the greens and browns 
outside to the bright blues and reds of the clothes you are wearing. All these colours result from the 
interaction of light with the pigments in these different things — some frequencies of light are 
absorbed, others scattered. Inside our eyes, chemical reactions detect these different frequencies and 
convert them into electrical nerve impulses sent to the brain. All these different pigments have one 
thing in common — lots of double bonds. For example, the pigment responsible for the red colour in 
tomatoes, lycopene, is a long-chain polyalkene. 




lycopene, the red pigment in tomatoes, rose hips, and other berries 

Lycopene contains only carbon and hydrogen while most pigments contain many other elements 
but nearly all contain double bonds. This chapter is about the properties, such as colour, of mole- 
cules that have several double bonds and that depend on the joining up or conjugation of the elec- 
trons in these double bonds. 

In earlier chapters, we talked about basic carbon skeletons made up of a bonds. In this chapter we 
shall see how, in some cases, we can also have a large Jt framework spread over many atoms and how 
this dominates the chemistry of such compounds. We shall see how this 71 framework is responsible 
for the otherwise unexpected stability of certain cyclic polyunsaturated compounds, including ben- 
zene and other aromatic compounds. We shall also see how this framework gives rise to the many 
colours in our world. To understand such molecules properly, we need to start with the simplest of 
all unsaturated compounds, ethene. 



The structure of ethene (ethylene, CH2=CH2) 

The structure of ethene (ethylene) is well known. It has been determined by electron diffraction and 
is planar {all atoms are in the same plane) with the bond lengths and angles shown below. The car- 
bon atoms are roughly trigonal and the C-C bond distance is shorter than that of a C-C bond. 



117.8 1 






C-H bond length 108 pm 
C=C bond length 133 pm 



152 



7 ■ Derealization and conjugation 



Important point. Ethene is not 
actually formed by bringing 
together two carbon atoms and 
four hydrogen atoms: individual 
carbon atoms do not hybridize their 
atomic orbitals and then combine. 
We are simply trying to rationalize 
the shapes of molecular orbitals. 
Hybridization and LCAO are tools to 
help us accomplish this. 



We shall use the approach of Chapter 4 (p. 106) and rationalize the shapes of molecular orbi- 
tals by combining the atomic orbitals of the atoms involved using the LCAO (Linear Combina- 
tion of Atomic Orbitals) approach. Hybridizing the atomic orbitals first makes this simpler. We 
mix the 2s orbital on each carbon atom with two of the three 2p orbitals to give three sp orbitals 
leaving the third p orbital unchanged. Two of the sp orbitals overlap with the hydrogen Is 
orbitals to form molecular orbitals, which will be the C-H bonds. The other sp orbital forms 
the C-C bond by overlapping with the sp orbital on the other carbon. The remaining p orbital 
can overlap with the p orbital on the other carbon to form a molecular orbital that represents the 
71 bond. 



hydrogen Is orbital (Q) 
carbon sp 2 orbital fl) 



carbon 2p orbital 
The two phases are 
shown in red and black 







we can think of the molecular orbitals as being 
a combination of the hybridized atomic orbitals 



C-H a-bonding orbital 



C-C ji-bonding orbital 
electron density above 
and below the a bond 
the different phases 
are shown in red and black 




C-C a-bonding orbital 

a simplified diagram of the 
bonding molecular orbitals of ethene 



a (i notation 



Theoretical chemists would label 
the energy of an electron in the p 
orbital as 'a' and that of an 
electron in the molecular orbital 
resulting from the combination of 
two p orbitals as 'a + P'. (Both a 
and p are, in fact, negative, which 
means that a + p is lower in energy 
than a.) The corresponding energy 
of an electron in the antibonding 
orbital is 'a-P'. Whilsta 
represents the energy an electron 
would have in an atomic orbital, p 
represents the change in energy 
when the electron is delocalized 
over the two carbon atoms. Since 
the % bond contains two electrons, 
both in the lowest-energy 
molecular orbital, then orbital, the 
total energy of the electrons is 2a 
+ 2p. If, instead, the two electrons 
remained in the atomic orbitals, 
their energy would be just 2a. 
Therefore the system is 2p lower in 
energy if the electrons are in then 
molecular orbital rather than the 
atomic orbitals. 



Q 

c c c-^*-c 

00 0^ 



Ethene is chemically more interesting than ethane because of the 7t bond. In fact, the % bond is the 
most important feature of ethene. In the words of Chapter 5, the C-C 7t orbital is the HOMO 
(Highest Occupied Molecular Orbital) of the alkene, which means that the electrons in it are more 
available than any others to react with something that wants electrons (an electrophile). Since this 
orbital is so important, we will look at it more closely. 

The 71 orbital results from combining the two 2p orbitals of the separate carbon atoms. 
Remember that when we combine two atomic orbitals we get two molecular orbitals. These result 
from combining the p orbitals either in-phase or out-of-phase. The in-phase combination accounts 
for the bonding molecular orbital (n), whilst the out-of-phase combination accounts for the anti- 
bonding molecular orbital (7t*). As we progress to compounds with more than one alkene, so the 
number of 71 orbitals will increase but will remain the same as the number of 7C* orbitals. 



SS 



c — c 

00 



2x C 




c 





b 



it* orbital higher in 
energy than carbon 
p orbital 



out-of-phase combination of 
p orbitals gives a it* orbital 



C3 



nodal plane between 
the two atoms 



it orbital lower in energy 
than carbon p orbital 



good overlap 



poor overlap 



the two p orbitals can only overlap 
if they are parallel 



in-phase combination of 
p orbitals gives a it orbital 



The 7C bond contains two electrons and, since we fill up the energy level diagram from the lowest- 
energy orbital upwards, both these electrons go into the bonding molecular orbital. In order to have 
a strong 7C bond, the two atomic p orbitals must be able to overlap effectively. This means they must 
be parallel. 



Molecules with more than one C-C double bond 



153 



There are two isomers ( cis and trans or E and Z) of many alkenes 

The 71 bond has electron density both above and below the bond as the parallel p orbitals overlap 
locking the bond rigid. Hence no rotation is possible about a double bond — the Tt bond must be bro- 
ken before rotation can occur. One consequence of this locking effect of the double bond is that there 
are two isomers of a disubstituted alkene. One is called a cis or Z alkene, the other a trans or E alkene. 



► 




By contrast, a C- 


■Co bond has 


electron density 


along the line 


joining the two nuclei and allows 


free rotation. 





Alkenes resist rotation 

Maleic and fumaric acids were known in the nineteenth 
century to have the same chemical composition and the 
same functional groups and yet they were different 
compounds — why remained a mystery. That is, until 1874 
when van't Hoff proposed that free rotation about double 
bonds was restricted. This meant that, whenever each 
carbon atom of a double bond had two different 




COOH 



HOOC ^H 

fumaric acid 



no change 



trans carboxyiic 
acid groups 



substituents, isomers would be possible. He proposed 
the terms cis (Latin meaning 'on this side') and trans 
(Latin meaning 'across or on the other side') for the two 
isomers. The problem was: which isomer was which? On 
heating, maleic acid readily loses waterto become maleic 
anhydride so this isomer must have both acid groups on 
the same side of the double bond. 



COOH 



COOH 




cis carboxyiic 
acid groups 



X 



a trans or £ alkene 



X 



a c/s or Z alkene 



maleic anhydride 



It is possible to interconvert cis and trans alkenes, but the TT bond must be broken first. This 
requires a considerable amount of energy — around 260 kj mol~ . One way to break the K bond 
would be to promote an electron from the % orbital to the %* orbital (from HOMO to LUMO). If 
this were to happen, there would be one electron in the bonding 7t orbital and one in the antibonding 
7t* orbital and hence no overall bonding. Electromagnetic radiation of the correct energy could pro- 
mote the electron from HOMO to LUMO. The correct energy actually corresponds to light in the 
ultraviolet (UV) region of the spectrum. Thus, shining UV light on an alkene would promote an 
electron from its bonding 7t molecular orbital to its antibonding %* molecular orbital, thereby 
breaking the 7t bond (but not the bond) and allowing rotation to occur. 



Q 



energy needed to 
promote electron 
from ji to 7t* 



c — c 



ftv 



i 



both electrons in it bond 



X 



no rotation is possible 



UV light breaks n bond 
rotation is now possible 



it* antibonding molecular orbital 



electron promoted to 
antibonding orbital 
cancels out electron 
in bonding orbital 
result: no it bond 



Tt bonding molecular orbital 



R R R H 



mixture of both isomers formed 



Notice that it takes less energy to 
break a C-C ji bond than a C-C o 
bond (about 260 kJ mol" 1 
compared to about 350 kJ mor 1 ). 
This is because the sideways 
overlap of the p orbitals to form an 
bond is not as effective as the 
head-on overlap of the orbitals to 
form a o bond. Enough energy is 
available to break the Tt bond if the 
alkene is heated to about 500 °C. 



00 



oc(Qco c c 



00 



overlap of orbitals is more efficient 
in a a bond than in a Tt bond 



Molecules with more than one C-C double bond 



Benzene has three strongly interacting double bonds 

The rest of this chapter concerns molecules with more than one C-C double bond and what happens 
to the Jt orbitals when they interact. To start, we shall take a bit of a jump and look at the structure of 
benzene. Benzene has been the subject of considerable controversy since its discovery in 1825. It was 



154 



7 ■ Derealization and conjugation 



soon worked out that the formula was CsHg, but how were these atoms arranged? Some strange 
structures were suggested until Kekule proposed the correct structure in 1865. 



these diagrams represent old structures 
for benzene. They do not represent 
compounds that could ever be made 



%^S /H 



o bonds shown in green 




a single p orbital 

different phases shown 

in red and black 






these early 
suggestions for the 
structure of benzene 
have now been made, 
^"they are certainly not 
benzene, but entirely 
different compounds 



H /%/ C \ H 



Kekule's structure for benzene 



Let's look at the molecular orbitals for Kekule's structure. As in simple alkenes, each of the carbon 
atoms is sp hybridized leaving the remaining p orbital free. 



The framework of the ben- 
zene ring is like the framework of 
an alkene. The problem comes 
with the p orbitals — which pairs 
do we combine to form the 71 
bonds? There seem to be two 
possibilities. 



comrjining different pairs ot p oroitais 
puts the double bonds in different postions 



X, 



> 



"X s* 





C0 2 H 



Combining these six atomic p orbitals 
actually produces six molecular 
orbitals. We shall consider the form of 
all these orbitals later in the chapter 
when we discuss benzene and 
aromaticity more fully. 




C0 2 H 




2-bromobenzoic acid '6'-bromobenzoic acid 
if the double bonds were localized then these two compound 
would be chemically different, (the double bonds are drawn 
shorter than the single bonds to emphasize the difference) 



With benzene itself, these two forms are 
equivalent but, if we had a 1,2- or a 1,3-dis- 
ubstituted benzene compound, these two 
forms would be different. A synthesis was 
designed for these two compounds but it was 
found that both compounds were identical. 
This posed a bit of a problem to Kekule — his 
structure didn't seem to work after all. His 
solution was that benzene rapidly equilibrates, 
or 'resonates' between the two forms to give 
an averaged structure in between the two. 

The molecular orbital answer to this problem, as you may well know, is that all six p orbitals can 
combine to form (six) new molecular orbitals, one of which (the one lowest in energy) consists of a 
ring of electron density above and below the plane of the molecule. Benzene does not resonate 
between the two Kekule structures — the electrons are in molecular orbitals spread equally over all 
the carbon atoms. However the term 'resonance' is still sometimes used (but not in this book) to 
describe this mixing of molecular orbitals. 

We shall describe the 71 electrons in benzene as 
delocalized, that is, no longer localized in specific 
double bonds between two particular carbon 

atoms but spread out, or delocalized, over all six v- -v the circle represents the 

atoms in the ring. An alternative drawing for delocalized system 

benzene shows the 71 system as a ring and does 
not put in the double bonds. 





Molecules with more than one C-C double bond 



155 



The Kekule structure is used for mechanisms 



This representation does present a slight problem 
to modern organic chemists, however: it is not 
possible to draw mechanisms using the 
delocalized representation of benzene. The curly 
arrows we use represent two electrons. This 
means that in order to write sensible mechanisms 
we still draw benzene as though the double bonds 
were localized. Keep in mind though that these 
double bonds are not really localized and it does 
not matter which way round we draw them. 




this representation 

cannot be used for 

mechanisms 



it does not matter 

which way round you draw 

the benzene ring - 

both drawings give the 

same result but one uses 

more arrows to get there 




We are saying that the K electrons are not localized in alternating double bonds but are actually 
spread out over the whole system in a molecular orbital shaped like a ring (we will look at the shapes of 
the others later). The electrons are therefore said to be delocalized. Theoretical calculations confirm 
this model, as do experimental observations. Electron diffraction studies show benzene to be a regu- 
lar, planar hexagon with all the carbon-carbon bond lengths identical (139.5 pm). This bond length 
is in between that of a carbon-carbon single bond (154.1 pm) and a full carbon-carbon double bond 
(133.7 pm). A further strong piece of evidence for this ring of electrons is revealed by proton NMR 
and is discussed on on p. 251. 

Derealization terminology 

What words should be used to describe delocalization is a vexed question. Terms 
such as resonance, mesomerism, conjugation, and delocalization are only a few of 
the ones you will find in books. You will already have noticed that we don't like 
'resonance' because it suggests that the structure vibrates rapidly between 
localized structures. We shall use conjugation and delocalization: conjugation 
focuses on the sequence of alternating double and single bonds while 
delocalization focuses on the molecular orbitals covering the whole system. 
Electrons are delocalized over the whole of a conjugated system. 

Noncyclic polyenes 

What would the structure be like if the 
three C-C double bonds were not in a ring 
as they are in benzene but were instead in a 
chain. What is the structure of hexatriene? 
Are the bond lengths still all the same? 

There are two isomers of hexatriene: a cis form and a trans form. The name refers to the 
geometry about the central double bond. The two isomers have different chemical and 
physical properties. Rotation is still possible about the single bonds (although slightly more 
difficult than around a normal single bond) and there are three different planar conformations 
possible for each isomer. Keeping the central black double bond the same, we can rotate about 
each of the green bonds in turn. Each row simply shows different ways to draw the same compound. 

\ 



cut here - 





benzene 



hexatriene 
a 'cut open benzene ring'? 




rotation about 



c/s-hexatriene 



single bond 




rotation about 



single bond 




rotation about 



single bond 




rotation about 



single bond 



frans-hexatriene 



156 



7 ■ Derealization and conjugation 



The structures of both cis- and frans-hexatriene have been determined by electron diffraction and 
two important features emerge. 

• Both structures are essentially planar (the cis form is not quite for steric reasons) 

• There are double and single bonds but the central double bond in each case is slightly longer than 
the end double bonds and the single bonds are slightly shorter than a 'standard' single bond 

this double bond is 137 pm 



This aspect of structure is called 
conformation and is the subject of 
Chapter 18. 




rotation about this bond 
requires only about 3 kJ mol -1 



because of the overlap of the 

p orbitals on carbons 2 and 3, 

rotation is now harder and requires 

more than 30 kJ mol -1 



i t 

H 



I 
H 



H 

I 



both single bonds are 146 pm 



both end double bonds are 134 pm 



standard values: single bond: 154 pm 
double bond: 134 pm 



The most stable structure of 
frans-hexatriene is shown 
here. 

Why is this structure planar 
and why are the bond lengths 
different from their 'standard' 
values? This sounds like the sit- 
uation with benzene and again 
the answers lie in the molecular 
orbitals that can arise from the 
combination of the six p 
orbitals. Just as in benzene, these orbitals can all combine to give one big molecular orbital over the 
whole molecule. However, the p orbitals can overlap and combine only if the molecule isplanar. 

Since the p orbitals on car- 
bons 2 and 3 overlap, there is 
some partial double bond 
character in the central 
bond, helping to keep the 
structure planar. This overlap 
means that it is slightly harder 
to rotate this 'formal single 
bond' than might be expect- 
ed — it requires about 30 
kJ mol~ to rotate it whereas 
the barrier in propene is only 
around 3 kJ mol - . 





when all the atoms are planar all six p orbitals can overlap 




\>-N* 




no overlap possible here 
if we rotate about a single bond, 
one pair of p orbitals can no longer overlap with the others 



This explains why the compound adopts a planar structure but, in order to understand why 
the bond lengths are slightly different from their expected values or even why they are not all 
the same as in benzene, we must look at the all the molecular orbitals for hexatriene. Before we 
can do this, we must first study some simpler systems and address the important question of conju- 
gation seriously. 



Conjugation 



In benzene and hexatriene every carbon atom is sp hybridized with the remaining p orbital available 
to overlap with its neighbours. The uninterrupted chain of p orbitals is a consequence of having 
alternate double and single bonds. When two double bonds are separated by just one single bond, 
the two double bonds are said to be conjugated. Conjugated double bonds have different properties 
from isolated double bonds, both physically (they are often longer as we have already seen) and 
chemically (Chapters 10, 23, and 35). 



Conjugated systems 

In the dictionary, 'conjugated' is defined, among other 

ways, as 'joined together, especially in pairs' and 'acting 

or operating as if joined'. This does indeed fit very well 

with the behaviour of such conjugated double bonds since describe electrons. 



the properties of a conjugated system are often different 
from those of the component parts. We are using 
conjugationto describe bonds and delocalizationto 



Conjugation 



157 




|3-carotene 



eleven double bonds are conjugated 



H 

propenal (acrolein): 

here the C-C double bond is 

conjugated with an aldehyde group 



You have already met several conjugated systems: 
remember lycopene at the start of this chapter and 
(3-carotene in Chapter 3? All eleven double bonds in 
(3-carotene are separated by only one single bond. We 
again have a long chain in which all the p orbitals can 
overlap to form molecular orbitals. 

It is not necessary to have two car- 
bon-carbon double bonds in order to 
have a conjugated system — the C-C 
and C-O double bonds of propenal 
(acrolein) are also conjugated. The 
chemistry of such conjugated carbonyl 
compounds is significantly different 
from the chemistry of their component 
parts (Chapter 10). 

What is important though is that the 
double bonds are separated by one and 
only one single bond. Remember the 
unsaturated fatty acid, linoleic acid, that you met in Chapter 3? Another fatty acid with even more 
unsaturation is arachidonic acid. None of the four double bonds in this structure are conjugated 
since in between any two double bonds there is an sp carbon. This means there is no p orbital avail- 
able to overlap with the ones from the double bonds. The saturated carbon atoms insulate the double 
bonds from each other. 



Me- 




-N ,-N- 


Me 

r 

y 


— Et 


Another very important 
highly conjugated 
compound is 
chlorophyll. This is the 
green pigment in plants 
without which life on 


Me" 


Mg' 

=n'' n- 


-Me 


earth as we know it 
could not exist. 


R0 2 C 


— -^ 


Me(K^\ 


^0 




the structure of chlorophyll 

the ring shown in green 

is fully conjugated 



these four double bonds are not conjugated 
- they are all separated by two single bonds 




these tetrahedral sp 3 carbons prevent any possible 
overlap of the p orbitals in the double bonds 

If an atom has two double bonds directly attached to it, that is, there are no single bonds separating "2^ C^CH 2 
them, again no conjugation is possible. The simplest compound with such an arrangement is allene. allene 

If we look at the arrangement of the p orbitals in this system, it is easy to see why no derealization 
is possible — the two 7t bonds are perpendicular to each other. 

central carbon is sp hybridized 
end carbons are sp 2 hybridized end carbons are sp 2 hybridized 




!iC :^C— C 

the % bonds formed as a result 

of the overlap of the p orbitals 

must be at right angles to each other 




not only are the two it bonds perpendicular, 
but the two methylene groups are too 



> Requirements for conjugation 

• Conjugation requires double bonds separated by one single bond 

• Separation by two single bonds or no single bonds will not do 



158 



7 ■ Derealization and conjugation 



butadiene 




The allyl system 

The allyl cation 

We would not say that two p orbitals are conjugated — they just , 

make up a double bond — so just how many p orbitals do we need 

before something can be described as conjugated? It should be 

clear that in butadiene the double bonds are conjugated — here ,, (~\ C~\ 

we have four p orbitals. -r ^^ \ I \ J \ I \J 

Is it possible to have three p orbitals interacting? How can we get an isolated p orbital-after all, we 
can't have half a double bond. Let us look for a moment at allyl bromide (prop-2-enyl bromide or 1- 
bromoprop-2-ene). Carbon 1 in this compound has got four atoms attached to it (a carbon, two 
hydrogens, and a bromine atom) so it is tetrahedral (or sp hybridized). 

Bromine is more electronegative than carbon and so the C-Br bond is polarized towards the 
bromine. If this bond were to break completely, the bromine would keep both electrons from the 
C-Br bond to become bromide ion, Br~, leaving behind an organic cation. The end carbon would 
now only have three groups attached and so it becomes trigonal (sp hybridized). This leaves a vacant 
p orbital that we can combine with the 71 bond to give a new molecular orbital for the allyl 




system. 



c (j ui ui lcii I la;. 



I ICLI y LU LUI 



ILHI IC Will I 



the % bond to form a new molecular orbital for the allyl system 




H 4r H 




As more orbitals combine it 
becomes more difficult to 
represent the molecular orbitals 
convincingly. We shall often, from 
now on, simply use the atomic 
orbitals to represent the 
molecular orbitals. 



► Remember: 

1 We are simply combining atomic orbitals here — whether or not any of 
the orbitals contain any electrons is irrelevant. We can simply fill up the 
resultant molecular orbitals later, starting with the orbitals lowest in 
energy and working upwards until we have used up any electrons we 
may have. 

2 This method allows us to work out, without too much difficulty, the 
shapes and energies of the molecular orbitals. The compound does 
not split its molecular orbitals into atomic orbitals and then 
recombine them into new molecular orbitals; we do. 



Rather than trying to combine the p orbital with the 71 bond, it is 
easier for us to consider how three p orbitals combine; after all, we 
thought of the 71 bond as a combination of two p orbitals. Since we 
are combining three atomic orbitals (the three 2p orbitals on car- 
bon) we shall get three molecular orbitals. The lowest-energy orbital 
will have them all combining in-phase. This is a bonding orbital 
since all the interactions are bonding. 

The next orbital requires one 
node, just as higher-energy atom- 
ic orbitals have extra nodes 
(Chapter 4). The only way to 
include a node and maintain the 
symmetry of the system is to put 
the node through the central 
atom. This means that when this 
orbital is occupied there will be 
no electron density on this cen- 



60a 

the bonding molecular 
orbital of the allyl system, m 1 

c .^-c 

om 

nodal plane 

through the middle atorr 

nonbonding *P 2 



tral atom. Since there are no 
interactions between adjacent atomic orbitals (either bonding or 
antibonding), this is a nonbonding orbital. 

The final molecular orbital must have two nodal planes. All the 
interactions of the atomic orbitals are out-of-phase so the resulting 
molecular orbital is an antibonding orbital. 







o 





two nodal planes 
antibonding m 3 



The allyl system 



159 



We can summarize all this information in a molecular orbital energy level diagram. 

the 7i molecular orbitals of the allyl system: the allyl cation 



3x 







'combine 



three degenerate 

2p orbitals combine 

to form three 

molecular orbitals 



-V3 



-y 2 



H 



-Vi 



antibonding 
molecular orbital 
higher in energy 
than a p orbital 



nonbonding 
molecular orbital 
Same energy 
as a p orbital 



bonding orbital, 
energy lower 
than p orbital 



three molecular orbitals 

resulting from the combinations 

of the three atomic orbitals 

the energies are now different 



this is the 
Lowest Unoccupied 
Molecular Orbital 
(LUMO) 



this is the 
Highest Occupied 
Molecular Orbital 
(HOMO) 



The two electrons that were in the % bond now occupy the orbital lowest in energy, the bonding 
molecular orbital VP^, and now spread over three carbon atoms. The electrons highest in energy and 
so most reactive are those in the HOMO. However, in this case, since the allyl cation has an overall 
positive charge, we wouldn't really expect it to act as a nucleophile. Of far more importance is the 
vacant nonbonding molecular orbital — the LUMO, the nonbonding ^2- K is this orbital that must 
be attacked if the allyl cation reacts with a nucleophile. From the shape of the orbital, we can see 
that the incoming electrons will attack the end carbon atoms not the middle one since, if this orbital 
were full, all electron density in it would be on the end carbon atoms, not the middle one. A different 
way of looking at this is to see which car- 
bon atoms in the system are most lacking 
in electron density. The only orbital in 
this case with any electrons in it is the 
bonding molecular orbital SPi. From the 
relative sizes of the coefficients on each 
atom we can see that the middle carbon 
has more electron density on it than the 
end ones; therefore the end carbons must 
be more positive than the middle one and 
so a nucleophile would attack the end 
carbons. 














060 



empty y 2 nonbonding MO 
if it did have electrons in it they 
would be on the end carbon atoms 
so nucleophiles attack here 



occupied \\i 1 bonding MO 
most electron density is on the 
central carbon which means the 
end carbons must be more 
positively charged 



Representations of the allyl cation 

How can we represent all this information with curly arrows? The simple answer is that we can't. 
Curly arrows show the movement of a pair of electrons. The electrons are not really moving around 
in this system — they are simply spread over all three carbon atoms with most electron density on the 
middle carbon. Curly arrows can give us an indication of the equivalence of the two end carbons, 
showing that the positive charge is shared over these two atoms. 

The curly arrows we used in this representation are slightly different from the curly arrows we 
used (Chapter 5) to represent mechanisms by the forming and breaking of bonds. We still arrive at 
the second structure by supposing that the curly arrows mean the movement of two electrons so that 
the right-hand structure results from the 'reaction' shown on the left-hand structure, but these 'reac- 
tions' would be the movement of electrons and nothing more. In particular, no atoms have moved 
and no O bonds have been formed or broken. These two structures are just two different ways of 



There are also all the molecular 
orbitals from the a framework but 
we do not need to consider these: 
the occupied cr-bonding molecular 
orbitals are considerably lower in 
energy than the molecular orbitals 
for the it system and the vacant 
antibonding molecular orbitals for 
the a bonds are much higher in 
energy than the it antibonding 
molecular orbital. 



The term coefficient describes 
the contribution of an individual 
atomic orbital to a molecular 
orbital. It is represented by the 
size of the lobes on each atom. 



curly arrows show the positive 

charge is shared over both the 

end atoms 



160 



7 ■ Derealization and conjugation 



drawing the same species. The arrows are derealization arrows and we use them to remind us that 
our simple fixed-bond structures do not tell the whole truth. To remind us that these are derealiza- 
tion arrows, we use a different reaction arrow, a single line with arrowheads on each end (<->). 

The problem with these structures is that they seem to imply that the positive charge (and the 
double bond for that matter) is jumping from one end of the molecule to the other. This, as we have 
seen, is just not so. Another and perhaps better picture uses dotted 
lines and partial charges. However, as in the representation of 
benzene with a circle in the middle, we cannot draw mechanisms 
on this structure. Each of the representations has its value and we 
shall use both. 



Do not confuse this derealization 
arrow with the equilibrium sign. A 
diagram like this would be wrong: 

The equilibrium arrows may be 
used only if atoms have moved 
and the species differ by at least 
a o bond. Maybe the simplest 
reaction that could be shown this 
way would be the protonation of 
water where a proton moves and 
an 0-H a bond, shown in black, is 
formed or broken. 



a structure to emphasize the 
equivalence of both bonds and the 
sharing of the charge at both ends 



H-% 



+ H 



H 

1® 



A summary of the allyl cation system 

• The two electrons in the 71 system are spread out over all three carbon atoms 
with most electron density on the central carbon 

• There are no localized double and single bonds — both C-C bonds are identi- 
cal and in between a double and single bond 

• Both end carbons are equivalent 

• The positive charge is shared equally over the two end carbons. The LUMO of 
the molecule shows us that this is the site for attack by a nucleophile 



The delocalized allyl cation can be compared to localized carbocations by NMR 

In the reaction below, a very strong acid (called 'superacid' — see Chapter 17) protonates the OH 
group of 3-cyclohexenol, which can then leave as water. The resulting cation is, not surprisingly, 
unstable and would normally react rapidly with a nucleophile. However, at low temperatures and if 
there are no nucleophiles present, the cation is relatively stable and it is even possible to record a car- 
bon NMR spectrum (at -80 °C). 

-oh „_. . rb H2 




FS0 3 H-SbF 5 



liquid S0 2 , -80 °C 





-H 2 





The NMR spectrum of this allylic cation reveals a plane of symmetry, which confirms that the 
positive charge is spread over two carbons. The large shift of 224 p.p.m. for these carbons indicates 
very strong deshielding (that is, lack of electrons) but is nowhere near as large as a localized cation. 
The middle carbon's shift of 142 p.p.m. is almost typical of a normal double bond indicating that it is 
neither significantly more nor less electron-rich than normal. 

141.9 
224.4 224.4 Carbocation 13 C shift 

This localized carbocation shows an enormous shift of 
330 p.p.m. indicating very little shielding of the positively 
charged carbon atom. Again, due to the instability of this 
species, the 13 C spectrum was recorded at low 



37.1 




temperature. 



37.1 



the 13 C NMR shifts in p.p.m. 
notice the plane of symmetry down the middle 



Me 



This carbon 
resonates 
'at 330 p.p.m. 




The allyl system 



161 




Br 




'Br 



H H 



homolytic cleavage of the C-Br bond forms 
a bromine atom and the allyl radical 



The allyl radical 

When we made the allyl cation from allyl bromide, the bromine atom 
left as bromide ion taking both the electrons from the C-Br bond with 
it — the C-Br bond broke heterolytically. What if the bond broke 
homolytically — that is, carbon and bromine each had one electron? A 
bromine atom and an allyl radical (remember a radical has an unpaired 
electron) would be formed. This reaction can be shown using the single- 
headed fish hook curly arrows from Chapter 5: normal double-headed 
arrows show the movement of two electrons; single-headed arrows show 
the movement of one. 

Now the end carbon has a single unpaired electron. What do we do with it? Before the bond 
broke, the end carbon was tetrahedral (sp hybridized). We might think that the single electron 
would still be in an sp orbital. However, since an sp orbital cannot overlap efficiently with a 7t 
bond, the single electron would then have to be localized on the end carbon atom. If the end carbon 
atom becomes trigonal (sp hybridized), the single electron could be in a p orbital and this could 
overlap and combine with the 71 bond. This would mean that the radical could be spread over the 
molecule in the same orbital that contained the cation. 

So once again we have three p orbitals to combine. This is the same situation as before. We 
have the same atoms, the same orbitals, and so the same energy levels. In fact, the molecular orbital 
energy level diagram for this compound is almost the same as the one for the allyl cation: the only 
difference is the number of electrons in the % system. Whereas in the allyl cation % system we only 
had two electrons, here we have three (two from the % bond plus the single one). Where does 
this extra electron go? Answer: in the next lowest molecular orbital — the nonbonding molecular 
orbital. 




inefficient overlap of 
sp 3 orbital and jt bond 




efficient overlap of 
p orbital and n bond 



■V3 



antibonding 
molecular orbital 
higher in energy 
than a p orbital 



3x 



'combine 



■V|/ 2 



nonbonding 
molecular orbital 
same energy as a 
p orbital 



this MO now has one 
electron in it. It is 
known as the Singly 
Occupied Molecular 
Orbital (SOMO) 
of the molecule 



■¥i 



bonding orbital 
energy lower than 
p orbital 



the n molecular orbitals of the allyl system: the allyl radical 

The extra electron is in an orbital all by itself. This orbital must be the HOMO of the molecule but 
is also the LUMO since it still has room for one more electron. It is actually called the Singly 

Occupied Molecular Orbital (SOMO), for 
obvious reasons. The shape of this orbital 
tells us that the single electron is located on 
the end carbon atoms. This can also be 
shown using derealization arrows (again 
single-headed arrows to show movement 
of one electron). 




H H H H 

the single electron can be on either of the end carbon atoms 



The allyl anion 

What would have happened if both electrons from the C-Br bond in allyl bromide had stayed behind 
on the carbon? If we had removed the bromine atom with a metal, magnesium for example (Chapter 
9), both electrons would remain leaving an overall negative charge on the allyl system. 



162 



7 ■ Derealization and conjugation 




-Y3 



-y 2 



this MO is now 
the LUMO 



this MO is now 
the HOMO 



0/H © 

+ MgBr 



reaction of allyl bromide with a metal gives the allyl anion 

Again, this system is much more 
stable if the negative charge can be 
spread out rather than localized on 
one end carbon. This can be accom- 
plished only if the negative charge is in 
a p orbital rather than an sp orbital. 
The molecular orbital energy level 
diagram is, of course, unchanged: all 
we have to do is put the extra electron 
in the nonbonding orbital. Altogether 
we now have four electrons in the 7C 
system — two from the 71 bond and 
two from the negative charge. Both 
the bonding and the nonbonding 
orbitals are now fully occupied. 
Where is the electron density in the allyl anion 71 system? The answer is slightly more complicated 
than that for the allyl cation because now we have two full molecular orbitals and the electron densi- 
ty comes from a sum of both orbitals. This means there is electron density on all three carbon atoms. 
However, the HOMO for the anion is now the nonbonding molecular orbital. It is this orbital that 
contains the electrons highest in energy and so most reactive. In this orbital there is no electron den- 
sity on the middle carbon; it is all on the end carbons. Hence it will be the end carbons that will react 
with electrophiles. This is conveniently represented by curly arrows. 



-Vi 



antibonding 
molecular orbital 



nonbonding 
molecular orbital 



bonding orbital 



the 7t molecular orbitals of the allyl system: the allyl anion 



*f*& 



^ 



the curly arrows give a good representation of the HOMO 

and they show the negative charge concentrated 

on the end carbon atoms. But the structures 

suggest localized bonds and charges 



these two structures emphasize 

the equivalence of the bonds and 

that the charge is spread out 



• A summary of the allyl anion system 

• There are no localized double and single bonds — both C-C bonds are the 
same and in between a double and single bond 

• Both end carbons are the same 

• The four electrons in the % system are spread out over all three carbon 
atoms. In the bonding orbital most electron density is on the central carbon 
but, in the nonbonding orbital, there is electron density only on the end car- 
bons 

• The electrons highest in energy and so most reactive (those in the HOMO) are 
to be found on the end carbons. Electrophiles will therefore react with the end 
carbons 

Such predictions from a consideration of the molecular orbitals are confirmed both by the reac- 
tions of the allyl anion and by its NMR spectrum. It is possible to record a carbon NMR spectrum of 
the allyl anion directly (for example, as its lithium derivative). The spectrum shows only two signals: 
the middle carbon at 147 p.p.m. and the two end carbons both at 51 p.p.m. 



Other allyl-like systems 



163 



The central carbon's shift of 147 p. p.m. is almost typical of a normal double bond carbon whilst 

the end carbons' shift is in between that of a double bond and a saturated carbon bearing a negative 

charge. Notice also that the central carbon in the allyl cation and the anion have almost identical 

chemical shifts — 142 and 147 p. p.m., respectively. If anything, the anion central carbon is more 

deshielded. Compare this with the spectra for methyllithium and propene itself. Methyllithium 

shows a single peak at -15 p. p.m. and propene shows three C signals as indicated below. 

methine carbon 
resonates at 134 p. p.m. 

H 
JO. 

CH^ CH 3 

* 4 



the methyl carbon 
resonates at -15 p. p.m. 



t 
CH 3 



-Li 



the central carbon 
resonates at 147 p. p.m. 



CHi 



H 





-CH, 



both end carbons 
resonate at 51 p. p.m. 



H 



CH^ 




CH 2 



a localized structure like this would 
have a very different spectrum 



methylene carbon 
resonates at 116 p. p.m. 



methyl carbon 
resonates at 19.5 p. p.m. 



Other allyl-like systems 

The carboxylate anion 

You may already be familiar with one anion very much like the allyl anion — the carboxylate ion 
formed on deprotonating a carboxylic acid with a base. In this structure we again have a double bond 
adjacent to a single bond but here oxygen atoms replace two of the carbon atoms. 



JU>" 



a carboxylic acid 



r» 



OH 



A. 



© 



R ^0 

a carboxylate anion 



H 2 



X-ray crystallography shows both carbon-oxygen bond lengths in this anion to be the same (136 
pm), in between that of a normal carbon-oxygen double bond (123 pm) and single bond (143 pm). 
The negative charge is spread out equally over the two oxygen atoms. 



5*o 



1 

^0 



le. 



© 



the electrons are delocalized over the n system 



OR 

'Of© 

these structures emphasize the equivalence 

of the two C-0 bonds and that the negative 

charge is spread over both oxygen atoms 



The molecular orbital energy diagram for the carboxylate anion is the very similar to that of the 
allyl system. There are just two main differences. 

1 The coefficients of the atomic orbitals making up the molecular orbitals will change because 
oxygen is more electronegative than carbon and so has a greater share of electrons 

2 The absolute values of the energy levels will be different from those in the allyl system, again 
because of the difference in the electronegativities. Compare with the differences between the 
molecular orbitals for ethene and a carbonyl, p. 103 



The nitro group 

The nitro group consists of a nitrogen bonded to two oxygen atoms and a carbon (for example, an 
alkyl group). There are two ways of representing the structure: one using formal charges, the other 
using a dative bond. Notice in each case that one oxygen is depicted as being doubly bonded, the 
other singly bonded. Drawing both oxygen atoms doubly bonded is incorrect — nitrogen cannot have 
five bonds since this would represent ten electrons around it and there are not enough orbitals to put 
them in. 



164 



7 ■ Derealization and conjugation 



Notice that the delocalization over 
the nitro group is similar to that 
over the carboxylate group. In 
fact, the nitro group is 
isoelectronic with the carboxylate 
group, that is, both systems have 
the same number of electrons. 



N© 









two ways of representing the nitro group 

the stucture on the left has formal charges 

on the nitrogen and one oxygen, the other 

has a dative bond from the nitrogen 



incorrect drawing 
of the nitro group 
nitrogen cannot 
have five bonds 



The problem with the two correct drawings is that they do not show the equivalence of the two 
N-O bonds. However, we do have an N-O double bond next to an N-O single bond which means 
that the negative charge is delocalized over both of the oxygen atoms. This can be shown by curly 
arrows. 






© 






2©0 



FT © 



OR 



'Oi© 



R © 



© 



the electrons are delocalized over the ji system 



these structures emphasize the equivalence 

of the two N-O bonds and that the negative 

charge is spread over both oxygen atoms 



y 



^Ts 



R 

an amide 



nitrogen is trigonal with 
its lone pair in a p orbital 



,H 



'(+) 




o^r 



Just to reiterate, the same molecular orbital energy diagram can be used for the allyl systems and 
the carboxylate and nitro groups. Only the absolute energies of the molecular orbitals are different 
since different elements with different electronegativities are used in each. 

The amide group 

The amide is a very important group in nature since it is the link by which amino acids join together 
to form peptides, which make up the proteins in our bodies. The structure of this deceptively simple 
group has an unexpected feature, which is responsible for much of the stability of proteins. 

In the allyl anion, carboxylate, and nitro systems we O 

had four electrons in the n system spread out over three 
atoms. The nitrogen in the amide group also has a pair of 
electrons that could conjugate with the 71 bond of the car- 

bonyl group. Again, for effective overlap with the 7C bond, ,. . . ,..,,«„, . 

' ° r ° r the lowest 7i orbital of the amide 

the lone pair of electrons must be in a p orbital. This in the same arrangement 

turn means that the nitrogen must be sp 2 hybridized. of p orbitals as in tne al| S" s y sten 

In the carboxylate ion, a negative charge was shared (equally) between two oxygen atoms. In an 
amide there is no charge as such — the lone pair on nitrogen is shared between the nitrogen and the 
oxygen. However, since oxygen is more electronegative ^ „ ©„ 

than nitrogen, it has more than its fair share of the elec- 
trons in this 7t system. (This is why the p orbital on the 
oxygen atom in the lowest bonding orbital shown above is 
slightly larger than the p orbital on the nitrogen.) The 
delocalization can be shown using curly arrows. 

This representation suffers from the usual problems. Curly arrows show the movement of a pair 
of electrons. The structure on the left, therefore, suggests that electrons are flowing from the nitrogen 
to the oxygen. This is not true: the molecular orbital picture tells us that the electrons are unevenly 
distributed over the three atoms in the % system with a greater electron density on the oxygen. The 
curly arrows show us how to draw an alternative diagram. The structure on the right implies that the 
nitrogen's lone pair electrons have moved completely on to the oxygen. Again this is not true; there is 
simply more electron density on the oxygen than on the nitrogen. The arrows are useful in that they 
help us to depict how the electrons are unevenly shared in the % system. 

A better representation might be this structure. The charges in brackets indicate substantial, 
though not complete, charges, maybe about a half plus or minus charge. However, we cannot draw 
mechanisms on this structure and all these representations have their uses. 



% 






.R ^ ^ 



© M 

N 



Other allyl-like systems 



165 



# Let us summarize these points. 

• The amide group is planar — this includes the first carbon atoms of the R 
groups attached to the carbonyl group and to the nitrogen atom 

• The lone pair electrons on nitrogen are delocalized into the carbonyl 
group 

• The C-N bond is strengthened by this interaction — it takes on partial 
double bond character. This also means that we no longer have free 
rotation about the C-N bond which we would expect if it were only a 
single bond 

• The oxygen is more electron-rich than the nitrogen. Hence we might expect 
the oxygen rather than the nitrogen to be the site of electrophilic attack 

• The amide group as a whole is made more stable as a result of the 
derealization 

The amide is a functional group of exceptional importance so we shall look at these points in 
more detail. 

The structure of the amide group 

How do we know the amide group is planar? X-ray crystal structures are the simplest answer. Other 
techniques such as electron diffraction also show that simple (noncrystalline) amides have planar 
structures. JV,N-dimethylformamide (DMF) is an example. 

The C-N bond length to the carbonyl group is closer to that of a standard C-N double bond (127 
pm) than to that of a single bond (149 pm). This partial double bond character is responsible for the 
restricted rotation about this C-N bond. We must supply 88 kj mol~ if we want to rotate the C-N 
bond in DMF (remember a full C-C double bond takes about 260 kj mol~ ). This amount of energy 
is not available at room temperature and so, for all intents 
and purposes, the amide C-N bond is locked at room tem- 
perature as if it were a double bond. This is shown in the 
carbon NMR spectrum of DMF. How many carbon signals 
would you expect to see? There are three carbon atoms 
altogether and three signals appear — the two methyl 
groups on the nitrogen are different. If free rotation were 
possible about the C-N bond, we would expect to see only 
two signals. In fact, if we record the spectrum at higher 
temperatures, we do indeed only see two signals since now 
there is sufficient energy available to overcome the rota- 
tional barrier and allow the two methyl groups to inter- 
change. 

Proteins are composed of many amino acids joined together with amide bonds. The amino group 
of one can combine with the carboxylic acid group of another to give an amide. This special amide, 
which results from the combining of two amino acids, is known as a peptide — two amino acids join 
to form a dipeptide; many join to give a polypeptide. 



X 



Me 



IT 

I 
Me 



DMF 

Di 

Methyl 

Formamide 



CH;, 



DMF al room temperature 



i * n ^\h ■* H i fh d t m B i m^gM aMyM w^ ^lNiMWtfirf^i^ MM qfw * fitinn0f0 9& i #^i 



200 



150 



H jH fr^ HMP ' mNWEM^ f* 



100 



At temperatures =■ ca. 150 °C 



50 



*4^WT4^^A^**^^ V l 4— K ' i »«l ■» * * ^ -l*'W|»' l f i«»— ' ' -, • * ?*• 



trK. 



*** W *«^*I* « -»»T' 



200 



150 



100 



50 



this special amide group 
is called the peptide unit 



H,N 




H,N 



OH 




H,0 



OH 



R 2 



amino acid 1 



amino acid 2 



H,N 





OH 



two amino acids, joined together by 
a peptide bond, form a dipeptide 



166 



7 ■ Derealization and conjugation 



The peptide unit so formed is a planar, rigid structure since there is restricted rotation about the 
C-N bond. This means that two isomers should be possible — a cis and a trans. 

It is found that nearly all the peptide 
units found in nature are trans. This is 



H 2 N 




H 

w 

H 2 N-{ 



C=0 and N-H are trans 



OH 



C=0 and N-H are cis 



not surprising since the cis form is more 
crowded (a trans disubstituted double 
bond is lower in energy than a cis for 
the same reason). 



Protein shape and activity 

This planar, trans peptide unit poses serious limitations 
on the shapes proteins can adopt. Understanding the 
shapes of proteins is very important — enzymes, for 



example, are proteins with catalytic properties. Their 
catalytic function depends on the shape adopted: alterthe 
shape in some way and the enzyme will no longer work. 



In Chapter 3 we saw that IR 
spectroscopy shows carbonyl 
groups in the region 1600-1800 
cm -1 . At the higher end of that 
region the C=0 stretches of very 
reactive acid chlorides and acid 
anhydrides show that they have 
full C=0 double bonds. At the 
lower end of that region the C=0 
stretching frequency of amides 
comes at about 1660 cm" 1 
showing that they are halfway to 
being single bonds. These 
relationships will be explored 
further in Chapter 15. The 
conjugation of the nitrogen's lone 
pair with the carbonyl bond 
strengthens the C-N bond but 
weakens the carbonyl bond. The 
weakerthe bond, the less energy 
it takes to stretch it and so the 
lower the IR absorption 
frequency. Overall the molecule 
is more stable, as is reflected in 
the reactivity (or lack of it) of the 
amide group, Chapter 12. 



% 



G 



3-^ ^N 



Reactivity of the amide group 

Just as derealization stabilizes the allyl cation, anion, and 
radical, so too is the amide group stabilized by the con- 
jugation of the nitrogen's lone pair with the carbonyl group. 
This, together with the fact that the amine part is such a 
poor leaving group, makes the amide one of the least reac- 
tive carbonyl groups (we shall discuss this in Chapter 12). 

Furthermore, the amine part of the amide group is unlike any normal amine group. Most amines 
are easily protonated. However, since the lone pair on the amide's nitrogen is tied up in the 7t system, 
it is less available for protonation or, indeed, reaction with any electrophile. As a result, an amide is 
preferentially protonated on the oxygen atom but it is difficult to protonate even there (see next 
chapter, p. 201). Conjugation affects reactivity. 



the oxygen atom's withdrawal of electrons 
weakens the carbonyl bond 



The conjugation of two n bonds 



The simplest compound that can have two conjugated 71 bonds is butadiene. As we would now 
expect, this is a planar compound that can adopt two different conformations by rotating about the 
single bond. Rotation is somewhat restricted (around 30 kj mol - ) but nowhere near as much as in 
an amide (typically 60-90 kj mol~ ). What do the molecular orbitals for the butadiene K system look 
like? The lowest-energy molecular orbital will have all the p orbitals combining in-phase. The next 
lowest will have one node, and then two, and the highest-energy molecular orbital will have three 
nodes (that is, all the p orbitals will be out-of-phase). 



Isomers of butadiene 

Butadiene normally refers to 1,3-butadiene. It is 
also possible to have 1,2-butadiene which is 
another example of an allene (p. 157). 



rotation about this single bond 
is only slightly restricted 



H '"«!C=C=C 



H 



\ 



CH, 



1,2-butadiene 
an allene 



H H - 

>=V' /" 

H >=C 

H H 

1,3-butadiene 
a conjugated diene 



The molecular orbitals of butadiene 

Butadiene has two Jt bonds and so four electrons in the Jt system. Which molecular orbitals are these 
electrons in? Since each molecular orbital can hold two electrons, only the two molecular orbitals 
lowest in energy are filled. Let's have a closer look at these orbitals. In 'Pj, the lowest-energy bonding 
orbital, the electrons are spread out over all four carbon atoms (above and below the plane) in one 
continuous orbital. There is bonding between all the atoms. The other two electrons are in W 2 . This 
orbital has bonding interactions between carbon atoms 1 and 2, and also between 3 and 4 but an 
antibonding interaction between carbons 2 and 3. Overall, in both the occupied % orbitals there are 



The conjugation of two jt bonds 



167 



electrons between carbons 1 and 2 and between 3 and 4, but the antibonding interaction between 
carbons 2 and 3 in *P 2 partially cancels out the bonding interaction in 'Pj. This explains why all the 
bonds in butadiene are not the same and why the middle bond is more like a single bond while the 
end bonds are double bonds. If we look closely at the coefficients on each atom in orbitals *Pi and 
y i?2> it can be seen that the bonding interaction between the central carbon atoms in 'Pi is greater 
than the antibonding one in *P2- Thus butadiene does have some double bond character between 
carbons 2 and 3, which explains why there is the slight barrier to rotation about this bond. 

antibonding 
antibonding interaction antibonding 
interaction . interaction 



\|/4 - 3 nodal planes 
bonding interactions 
3 antibonding interactions 
overall - antibonding orbital 



H H 




H H 

butadiene 




antibonding antibonding 
interaction interaction 




bonding 
interaction 



antibonding 
interaction 




bonding 
interaction 




bonding 



bonding 



interaction bonding interaction 
interaction 



\|<3 - 2 nodal planes 

1 bonding interaction 

2 antibonding interactions 
overall - antibonding orbital 



\|/ 2 - 1 nodal plane 
2 bonding interactions 
1 antibonding interaction 
overall - bonding orbital 



yi - nodal planes 
3 bonding interactions 
antibonding interactions 
overall - bonding orbital 



In our glimpse of hexatriene earlier in this chapter we saw a similar effect, which we could now 
interpret if we looked at all the molecular orbitals for hexatriene. We have three double bonds and 
two single bonds with slightly restricted rotation. Both butadiene and hexatriene have double bonds 
and single bond: neither compound has all its C-C bond lengths the same, yet both compounds are 
conjugated. What is the real evidence for conjugation? How does the conjugation show itself in the 
properties and reactions of these compounds? To answer these questions, we need to look again at 
the energy level diagram for butadiene and compare it with that of ethene. A simple way to do this is 
to make the orbitals of butadiene by combining the orbitals of ethene. 



168 



7 ■ Derealization and conjugation 



Recall that on p. 153 we saw how it 
was possible to promote an electron 
from HOMO to LUMO in an isolated 
double bond using UV light and that 
this allowed rotation about this bond. 
In butadiene, however, promoting an 
electron from the HOMO to the LUMO 
actually increases the electron density 
between the two central atoms and so 
stops rotation. 




¥4 



00 
0D 



00 

o 



combine 
out-of-phase 



combine 
in-phase 




¥3 LUMO 



a 




ft 



¥2 HOMO 



ii Oh 



combine 
out-of-phase 

combine 
in-phase 




ft 



¥l 



We have drawn the molecular orbital diagram for the 71 molecular orbitals of butadiene as a result 
of combining the 71 molecular orbitals of two ethene molecules. There are some important points to 
notice here. 

• The overall energy of the two bonding butadiene molecular orbitals is lower than that of the two 
molecular orbitals for ethene. This means that butadiene is more thermodynamically stable than 
we might expect if its structure were just two isolated double bonds 

• The HOMO for butadiene is higher in energy relative to the HOMO for ethene. This means buta- 
diene should be more reactive than ethene towards nucleophiles 

• The LUMO for butadiene is lower in energy than the LUMO for ethene. Consequently, butadiene 
would be expected to be more reactive towards nucleophiles than ethene 

• So whilst butadiene is more stable than two isolated double bonds, it is also more reactive (Chapter 20) 



Butadiene model 

A simple theoretical model of the butadiene system 
predicts the energy of the bonding "Pj orbital to be [a + 
1.62p] and that of bonding orbital <P 2 10 be [« + 0.620]. 
With both of these orbitals fully occupied, the total energy 
of the electrons is [4a + 4.48p[. Remember that the 
energy of the bonding 7t molecular orbital for ethene was 
[a + (3] (p. 152) so, if we were to have two localized ji 



bonds (each with two electrons), the total energy would be 
[4a + 4p]. This theory predicts that butadiene with both 
double bonds conjugated is lower in energy than it would 
be with two localized double bonds by 0.48fS. Both a and (3 
are negative; hence [4a + 4.48|3] is lower'm energy than 
[4a + 4p] by 0.48p\ 



UV and visible spectra 



169 



UV and visible spectra 



In Chapter 2 we saw how, if given the right amount of energy, electrons can be promoted from a low- 
energy atomic orbital to a higher-energy one and how this gives rise to an atomic absorption spec- 
trum. Exactly the same process can occur with molecular orbitals. In fact, we have already seen 
(p. 153) that UV light can promote an electron from the HOMO to the LUMO in a double bond. 



HOMO-LUMO gap 

Electrons can be promoted from any filled orbital to any 
empty orbital. The smallest energy difference between a 
full and empty molecular orbital is between the HOMO and 
the LUMO. The smallerthis difference, the less energy will 
be needed to promote an electron from the HOMO to the 
LUMO: the smaller the amount of energy needed, the 
longer the wavelength of light needed since AE= hv. 
Therefore, an important measurement is the wavelength 



at which a compound shows maximum absorbance, ^. max . 
A difference of more than about 4 eV (about 7 x 10~ 19 J) 
between HOMO and LUMO means that X max will be in the 
ultraviolet region (wavelength, X, < 300 nm). If the energy 
difference is between about 3 eV (about 4 x 10~ 19 J) and 
1.5eV(about3xlO~ 19 J)then^. max will be in the visible 
part of the spectrum. 



We can get a good estimate of the 
absolute energies of molecular orbitals 
from photoelectron spectroscopy and 
electron transmission spectroscopy 
(see Chapter 2). Such experiments 
suggest energies for the HOMO and 
LUMO of butadiene to be -9.03 and 
+0.62 eV, respectively, whilst for 
ethene they are -10.51 and +1.78 eV, 
respectively. 



We have seen above that the energy difference between the HOMO and LUMO for butadiene is 
less than that for ethene. Therefore we would expect butadiene to absorb light of longer wavelength 
than ethene (the longer the wavelength the lower the energy, AE= hc/X). This is found to be the case: 
butadiene absorbs at 215 nm compared to 185 nm for ethene. The conjugation in butadiene means it 
absorbs light of a longer wavelength than ethene. In fact, this is true generally. 

The more conjugated a compound is, the smaller the energy transition between its 
HOMO and LUMO and hence the longer the wavelength of light it can absorb. Hence 
UV-visible spectroscopy can tell us about the conjugation present in a molecule. 



orbitals 
of ethene 



LUMO 71*' 



HOMO to LUMO 
excitation: 
large gap: 
absorption 
in far UV 
at 185 nm 



HOMO 



"# 



orbitals 
of butadiene 



¥4 



¥3 LUMO 



HOMO to LUMO 
excitation: 
smaller gap: 
absorption 
in nearer UV 
at 215 nm 



ff 



¥2 HOMO 



# 



¥l 



170 



7 ■ Derealization and conjugation 



Both ethene and butadiene absorb in the far-UV region of the electromagnetic spectrum (215 nm is 
just creeping into the UV region) but, if we extend the conjugation further, the gap between HOMO and 
LUMO will eventually be sufficiently decreased to allow the compound to absorb visible light and hence 
be coloured. A good example is the red pigment in tomatoes we introduced at the start of the chapter. It 
has eleven conjugated double bonds (plus two unconjugated) and absorbs light at about 470 nm. 




lycopene, the red pigment in tomatoes, rose hips, and other berries 



The colour of pigments depends on conjugation 

You can see now that it is no coincidence that this compound and the two other highly conjugated 
compounds we met earlier, chlorophyll and (3-carotene, are all highly coloured natural pigments. In 
fact, all dyes and pigments are highly conjugated compounds. 

Natural pigments 

The similarities between 
lycopene and p-carotene 
are easierto see if the 
structure of lycopene is 
twisted. Lycopene is a 
precursor of carotene 
so, when a cell makes 
carotene, it makes 
lycopene en route. 




lycopene, the red pigment in tomatoes, rase hips, and other berries 




P-carotene, the red pigment in carrots and other vegetables 

If a compound absorbs one colour, it is the complementary colour that is transmitted — the red 
glass of a red light bulb doesn't absorb red light; it absorbs everything else letting only red light 
through. Here is a table of approximate wavelengths for the various colours. The last column gives 
the approximate length a conjugated chain must be in order to show the colour in question. The 
number n refers to the number of double bonds in conjugation. 



Approximate wavelengths for different colours 

Absorbed frequency, nm Colour absorbed Colour transmitted 

200-400 ultraviolet — 

400 
425 
450 
490 
510 
530 
550 
590 
640 
730 



violet 


yellow-green 


indigo-blue 


yellow 


blue 


orange 


blue-green 


red 


green 


purple 


yellow-green 


violet 


yellow 


indigo-blue 


orange 


blue 


red 


blue-green 


purple 


green 



R(CH=CH)„R, „ : 

< 8 

8 
9 
10 

11 



Every extra conjugated double bond in a system increases the wavelength of light that is 
absorbed. If there are fewer than about eight conjugated double bonds, the compound absorbs in 



Aromaticity 



171 



the ultraviolet and we don't notice the difference. With more than eight conjugated double bonds, 
the absorption creeps into the visible and, by the time it reaches 11, the compound is red. If we want- 
ed a blue or green compound, we should need a very large number of conjugated double bonds and 
such pigments do not usually rely on Jt bonds alone. 

Transitions from bonding to antibonding Jt orbitals are called 7t — > 7t* transitions. A much small- 
er energy gap is available if we use electrons in a nonbonding orbital as the electrons start off much 
higher in energy and can be promoted to low-lying antibonding 71 orbitals. We call these transitions 
n — » Jt*, where the 'n' stands for nonbonding. It is easy to find coloured compounds throughout the 
whole range of wavelengths by this means. The colour of blue jeans comes from the pigment indigo. 
The two nitrogen atoms provide the lone pairs that can be excited into the 71* orbitals of the rest of 
the molecule. These are low in energy because of the two carbonyl groups. Yellow light is absorbed by 
this pigment and indigo-blue light transmitted. 







colourless precursor to indigo 



indigo: the pigment of blue jeans 





N0 2 



Jeans are dyed by immersion in a vat of reduced indigo, which is colourless since the conjugation 
is interrupted by the central single bond. When the cloth is hung up to dry, the oxygen in the air oxi- 
dizes the 'pigment' to indigo and the jeans turn blue. Conjugation is the key to colour. 

Many conjugated compounds are yellow because, 
although they have their \ max in the UV, the broad 
absorption tails into the visible and the compound weakly 
absorbs violet light making it pale yellow. An example is 
this imine with a long conjugated system joining the two 
aromatic rings together. 
The imine is yellow but when it is reduced to the amine, breaking the conjugation in the middle so 
that the two benzene rings are no longer linked together, the result is a dark orange compound. This 
is rather surprising because you would normally expect the compound with the longer conjugated 
system to absorb at longer wavelengths. Check with the table above to see that an orange compound 
definitely absorbs at longer wavelengths than a yellow compound. 



% 





NO, 



NaBH 4 



the yellow imine 




the orange amine 




NO, 



The answer to this paradox lies in the change of hybridization of the nitrogen atom. In the imine, 
the nitrogen is trigonal and the lone pair is in an sp orbital in the plane of the conjugated system. No 
derealization of the lone pair is possible and the UV absorption comes from a simple 7C — > 7t* tran- 
sition. When the imine is reduced, the C-N bond can rotate and the amine can be trigonal too, but 
with the N-H bond in the plane and the lone pair in a p orbital conjugated with the right-hand ben- 
zene ring. The absorption giving the orange colour is an n — > 71* transition not a % — » Jt* transition. 
Even derealization of a lone pair into one benzene ring with a nitro group can give a longer wave- 
length absorption than a conjugated system of bonding electrons. 



Aromaticity 



Let us now return to the structure of benzene. Benzene is unusually stable for an alkene and is not 
normally described as an alkene at all. For example, whereas normal alkenes readily react with 



172 



7 ■ Derealization and conjugation 



bromine to give dibromoalkane addition products, benzene reacts with bromine only with difficul- 
ty — it needs a Lewis acid catalyst and then the product is a mono substituted benzene and not an 



This chemistry is discussed in Chapter 
22. 




cyclooctatetraene 



X 



alkene 



Br? 



R | / Br 

B 7x« 



dibromo 
addition 
product 
formed 





Br 2 


H 

1 


\/ , 


A 


w 




H^^V^H 




H 


FeBr 3 / Br 2 








monosubstituted 

product 

formed 



addition compound. 

Bromine reacts with ben- 
zene in a substitution reac- 
tion (a bromine atom 
replaces a hydrogen atom), 
keeping the benzene structure 
intact. This ability to retain 
its ring structure through all 
sorts of chemical reactions is 
one of the important differ- 
ences of benzene compared 
to alkenes and one that origi- 
nally helped to define the 
class of aromatic compounds 
to which benzene belongs. 

Cyclooctatetraene has 
four double bonds in a ring. 
What do you think its struc- 
ture will be? 

You will probably be surprised to find cyclooctatetraene (COT for short), unlike benzene, is not 
planar. Also none of the double bonds are conjugated — there are indeed alternate double and single 
bonds in the structure but conjugation is possible only if the p orbitals of the double bonds can over- 
lap; here they do not. Since there is no conjugation, there are two C-C bond lengths in cyclooctate- 
traene — 146.2 and 133.4 pm — which are typical for single and double C-C bonds. If possible, make 
a model of cyclooctatetraene for yourself — you will find the compound naturally adopts the shape 
below. This shape is often called a 'tub'. 

Chemically, cyclooctatetraene behaves like an alkene not 
like benzene. Bromine, for example, does not form a substi- 
tution product but an addition product. There is something 
strange going on here — why is benzene so different from 
other alkenes and why is cyclooctatetraene so different from 
benzene? The mystery deepens when we look at what happens when we treat cyclooctatetraene with 
powerful oxidizing or reducing agents. 

If 1,3,5,7-tetramethylcyclooctatetraene is treated at low temperature (-78 °C) with SbF 5 /S02ClF 
(strongly oxidizing conditions) a dication is formed. This cation, unlike the neutral compound, is 
planar and all the C-C bond lengths are the same. 

Me 

Me 




Me 



Me 





SbF 5 / S0 2 CIF 



-78 °C 



H Me 

neutral compound is tub-shaped 




Me 

dication is planar 



Drawing the dication 

The dication still has the same number of atoms as the neutral 
species only fewer electrons. Where have the electrons been 
taken from? The n system now has two electrons less. We 
could draw a structure showing two localized positive charges 
but this would not be ideal since the charge is spread over the 
whole ring system. 




one structure with 
localized charges 






the charges can be delocalized 
all round the ring 



structure to show equivalence 
of all the carbon atoms 



Aromaticity 



173 



SiMe 3 




SiMe 3 



SiMe 3 



SiMe 3 



Li / THF 



Me 



Me 



SiMe 3 

planar neutral compound 




"SiMe 3 
SiMe 3 

nonplanar dianion 



It is also possible to add electrons to cyclooctatetraene by treating it with alkali metals and a dian- 
ion results. X-ray structures reveal this dianion to be planar, again with all C-C bond lengths the 
same (140.7 pm). The difference between the anion and cation of cyclooctatetraene on the one hand 
and cyclooctatetraene on the other is the number of electrons in the 71 system. The cation has six 71 
electrons, the anion has ten, but neutral cyclooctatetraene has eight. 

Substituted benzene compounds, such as the one below 
with six silicon atoms around the edge, can also react with 
lithium to give a dianion. This dianion, with eight 71 elec- 
trons, is now no longer planar. 

Treatment of benzene itself with the strongly oxidizing 
SbF 5 /S02ClF reagent has no effect but it is possible to oxi- 
dize substituted derivatives. Hexakis(dimethylamino)- 
benzene, for example, can be oxidized with iodine. Again, 
the resulting dication is nonplanar and all the C-C bond 
lengths are not the same. 

Do you see a pattern forming? The important point is 
not the number of conjugated atoms but the number of 
electrons in the K system. When they have 4 or 8 7t electrons, 
both benzene and cyclooctatetraene adopt nonplanar 
structures; when they have 6 or 10 7C electrons, a planar 
structure is preferred. 

If you made a model of cyclooctatetraene, you might have tried to force it to be flat. If you man- 
aged this you probably found that it didn't stay like this for long and that it popped back into the tub 
shape. The strain in planar COT can be overcome by the molecule adopting the tub conformation. 
The strain is due to the numbers of atoms and double bonds in the ring — it has nothing to do with 
the number of electrons. The planar dication and dianion of COT still have this strain. The fact that 
these ions do adopt planar structures must mean there is some other form of stabilization that out- 
weighs the strain of being planar. This extra stabilization is called aromaticity. 



NMe 2 



Me 2 N 




NMe, 



Me 2 IT y ^NMe 2 

NMe 2 

planar neutral compound 



Heats of hydrogenation of benzene and cyclooctatetraene 

It is possible to reduce unsaturated C=C double bonds using hydrogen gas and a catalyst (usually nickel 
or palladium) to produce fully saturated alkanes. This process is called hydrogenation and it is exother- 
mic (that is, energy is released) since a thermodynamically more stable product, an alkane, is produced. 



Me,N 



Me 2 N 




NMe 2 

nonplanar dication 



Margarine manufacture 

This reaction is put to good use in the manufacture of 
margarines. One of the ingredients in many margarines is 
hydrogenated vegetable oil. When polyunsaturated fats 
are hydrogenated they become more solid. This means 



H 2 / Ni 



H 2 / Ni 



that, rather than having to pour our margarine on to our 
toast in the morning, we can spread it. We saw the second 
acid in this series, linoleic acid, at the start of Chapter 2. 



^C0 2 H 



„C0 2 H 



,C0 2 H 



H 2 / Ni 
r 

,C0 2 H 

melting points (m.p.s) of some common fatty acids 



linolenic acid 
m.p. -11 °C 



linoleic acid 
m.p. -5° C 



oleic acid 
m.p. 16 °C 



stearic acid 
m.p. 71 °C 



When ds-cyclooctene is hydrogenated, 96 kj mol of energy is released. Cyclooctatetraene 
releases 410 kj mol - on hydrogenation. This value is approximately four times one double 



174 



7 ■ Derealization and conjugation 



bond's worth, as we might expect. However, whereas the heat of hydrogenation for cyclohexene 
is 120 kj mol , on hydrogenating benzene, only 208 kj mol is given out, which is much less than 
the 360 kj mol~ that we would have predicted. This is shown in the energy level diagram below. 




I] +4H 2 




H 2+ Q 



Benzene has six % molecular orbitals 

The difference between the amount of energy we expect to get out on hydrogenation (360 kj mol~ ) 
and what is observed (208 kj mol~ ) is about 150 kj mol~ . This represents a crude measure of just 
how extra stable benzene really is relative to what it would be like with three localized double bonds. 
In order to understand the origin of this stabilization, we 
must look at the molecular orbitals. We can think of the 7t 
molecular orbitals of benzene as resulting from the combina- 
tion of the six p orbitals. We have already encountered the 
molecular orbital lowest in energy with all the orbitals com- 
bining in-phase. 

The next lowest molecular orbital will have one nodal plane. How can we divide up the six atoms 
symmetrically with one nodal plane? There are two ways depending on whether or not the nodal 
plane passes through a bond or an atom. 





the lowest energy MO for benzene has 
all the p orbitals combining in-phase 



m 




and 



"V 




ncfdal plane through atoms 



nodal plane through bonds 



there are two ways of symmetrically dividing the six carbon atom 
one has a node through two atoms, the other through two C-C bonds 

It turns out that these two different molecular orbitals both have exactly the same energy, that is, 
they are degenerate. This isn't obvious from looking at them but, nevertheless, it is so. 

The next molecular orbital will have two nodal planes and again there are two ways of arranging 
these, which lead to two degenerate molecular orbitals. 



Aromaticity 



175 







and 





with two nodal planes, there are again two possible molecular orbitals 







the MO highest in energy has all p orbitals combining out-of-phase 



Benzene model 

Whilst the HOMOs for benzene are the degenerate it 
molecular orbitals ("Pa), the next molecular orbital 
down in energy is not actually the n molecular orbital 
OPi). This all-bonding n molecular orbital OPi) is so 
stable that foura bonding molecular orbitals actually 
come in between the it molecular *Pi and ¥2 orbitals 
but are not shown in this molecular orbital energy level 
diagram. The greatest contribution to stability comes 
from this lowest-energy it bonding molecular orbital 
("Pi). This allows bonding interactions between all 
adjacent atoms. Theory tells us that the energy of this 
orbital is a + 2(S whilst that of the degenerate bonding 
molecular orbitals is a + (5. When all these bonding 
molecular orbitals are fully occupied, the total energy 
of the electrons is 6a + 8fS, which is 2(3 lower in energy 
than we would predict forthree localized double 
bonds. Butadiene had a theoretical stabilization 
energy of just 0.48(3 relative to two isolated double 
bonds so 2(3 is really quite significant. 



The final molecular orbital will have 
three nodal planes, which must mean all 
the p orbitals combining out-of-phase. 

These then are the six 71 molecular 
orbitals for benzene. We can draw an ener- 
gy level diagram to represent them. 




LUMO 




¥3 



¥3 



antibonding molecular orbitals 



LUMO 




HOMO <^?~ V-> -\\T V2 V2 -ft- VY YA HOMO 




bonding molecular orbitals 



the it molecular orbitals for benzene. The dashed line represents the energy of an isolated p orbital 
all orbitals below this line are bonding, all above it are antibonding. 
Benzene has six electrons in its it system so all the bonding MOs are fully occupied 



The 71 molecular orbitals of conjugated cyclic hydrocarbons can be easily predicted 

Notice that the layout of the energy levels is a regular hexagon with its apex pointing downwards. It 
turns out that the energy level diagram for the molecular orbitals resulting from the combination of 
any regular cyclic arrangement of p orbitals can be deduced from the appropriately sided polygon. If 
we take a regular polygon with one corner pointing downwards and draw a circle round it so that all 
the corners touch the circle, the energies of the molecular orbitals will be where the corners touch the 
circle. The circles should be of the same size and the polygons fitted inside the circle. The horizontal 
diameter represents the energy of a carbon p orbital and so, if any energy levels are on this line, they 
must be nonbonding. All those below are bonding; all those above antibonding. 




n = 6 n = b n = 4 n- 

^~ = energy level of MO relative to energy of p orbital indicated by dashed line 
n = number of carbon atoms in ring 



176 



7 ■ Derealization and conjugation 



Notes on these energy level diagrams: 

• This method predicts the energy levels for the molecular orbitals of planar, monocyclic, arrange- 
ments of identical atoms (usually all C) only 

• The dashed line represents an energy level a and in each case the circle radius is 2(3 

• There is always one single molecular orbital lower in energy than all the others (at energy a + 2(3). 
This is because there is always one molecular orbital where all the p orbitals combine in-phase 

• If there are an even number of atoms, there is also a single molecular orbital highest in energy; 
otherwise there will be a pair of degenerate molecular orbitals highest in energy 

• All the molecular orbitals come in degenerate pairs except the one lowest in energy and, for even- 
numbered systems, the one highest in energy 

Now we can begin to put all the pieces together and make sense of what we know so far. Let us 
compare the energy level diagrams for benzene and planar cyclooctatetraene. We are not concerned 
with the actual shapes of the mole- ^™ ^™ 

cular orbitals involved, just the 
energies of them. 

Benzene has six % electrons, 
which means that all its bonding 
molecular orbitals are fully occu- 
pied giving a closed shell structure. 
COT, on the other hand, has eight 
electrons. Six of these fill up the 
bonding molecular orbitals but there are two electrons left. These must go into the degenerate pair of 
nonbonding orbitals. Hund's rule (Chapter 4) would suggest one in each. Therefore this planar struc- 
ture for COT would not have the closed shell structure that benzene has — it must either lose or gain 
two electrons in order to have a closed shell structure with all the electrons in bonding orbitals. This is 
exactly what we have already seen — both the dianion and dication are planar, allowing derealization 
all over the ring, whereas neutral COT adopts a nonplanar tub shape with localized bonds. 



Of course, this isn't the molecular 
orbital energy level diagram for 
real cyclooctatetraene since COT 
is not planar but tub-shaped. 



+ + + 

MO level diagram for benzene 



+ 



f 



t 






MO level diagram for planar 
cyclooctatetraene 



This is not a strict definition of 
aromaticity. It is actually very difficult to 
give a concise definition. Huckel's rule 
is certainly a good guide but also 
important is the extra stability of the 
compound (shown, for example, in 
resistance to changes to its ji system) 
and low reactivity towards 
electrophiles. Perhaps the best 
indication as to whether or not a 
compound is aromatic is the proton 
NMR spectrum. The protons attached 
to an aromatic ring are further 
downfield than would otherwise be 
expected (Chapter 11). 



Annulenes (meaning ring 
alkenes) are compounds with 
alternating double and single 
bonds. The number in brackets 
tells us how many carbon atoms 
there are in the ring. Using this 
nomenclature, you could call 
benzene [6]annulene and 
cyclooctatetraene [8]annulene — 
but don't. 



Hiickel's rule tells us if compounds are aromatic 

Using this simple method to work out the energy level diagrams for other rings, we find that there is 
always a single low-energy bonding orbital (composed of all p orbitals combining in-phase) and then 
pairs of degenerate orbitals. Since the single orbital will hold two electrons when full and the degen- 
erate pairs four, we shall have a closed shell of electrons in these it orbitals only when they contain 2 
+ An electrons (« is an integer 0, 1, 2, etc.). This is the basis of Hiickel's rule. 

Hiickel's rule 

Planar, fully conjugated, monocyclic systems with (4n + 2) Jt electrons have a 
closed shell of electrons all in bonding orbitals and are exceptionally stable. Such 
systems are said to be aromatic. 

Analogous systems with An K electrons are described as anti-aromatic 

That the Jt system is fully conjugated and planar are important conditions for 
aromaticity. The next (An + 2) number after six is ten so we might expect this 
cyclic alkene to be aromatic. 

If this annulene with five cis double bonds were pla- 
nar, each internal angle would be 144°. Since a normal 
double bond has bond angles of 120°, this would be far 
from ideal. This compound can be made but it does not adopt a planar con- 
formation and therefore is not aromatic even though it has ten it electrons. 
By contrast, [18] annulene, which is also a (An + 2) it electron system (n = 4), 
does adopt a planar conformation and is aromatic (as shown by proton [l8]-annulene 




all-c/s-[10]annulene 




Aromaticity 



177 



NMR). Note the trans-trans-cis double bonds: all bond angles can be 120°. [20] annulene presumably 
could become planar (it isn't quite) but since it is a 4n p electron system rather than a 4n + 2 system, 
it is not aromatic and the structure shows localized single and double bonds. 

The importance of the system being monocyclic is less clear. The problem that often arises is 
'exactly how do we count the 71 electrons?'. Taking a simple example, should we consider naphtha- 
lene as two benzene rings joined together or as a ten 71 electron system? 

1,6-Methano[10]annulene is rather like 
naphthalene but with the middle bond 
replaced by a methylene bridging group. 
This compound is almost flat (carbons 
1 and 6 are raised slightly out of the 
plane) and shows aromatic character. 





should we count naphthalene as two benzene 
rings or one large ring with 10 % electrons? 




From its chemistry, it is very clear that naphthalene is aromatic but perhaps a little less so than 
benzene itself. For example, naphthalene can easily be reduced to tetralin (1,2,3,4-tetrahydronaph- 
thalene) which still contains a benzene ring. Also, in contrast to benzene, all the bond lengths in 
naphthalene are not the same. 




Na / ROH 



heat 




naphthalene 



tetralin 



137 pm-._ 



140 pm- 



142 pm 




133 pm 



naphthalene 



o<T s - 




© 



BH 



deprotonation of cyclopentadiene gives 
the stable cyclopentadienyl anion 



Htickel's rule is very useful and it helps us to predict and understand the aromatic stability of 
numerous other systems. Cyclopentadiene, for example, has two double bonds that are conjugated 
but the whole ring is not conjugated since there is a methylene group in the ring. However, this com- 
pound is relatively easy to deprotonate (see next chapter, p. 000) to give a very stable anion in which 
all the bond lengths are the same. How many 
electrons does this system have? Each of the 
double bonds contributes two electrons and the 
negative charge (which must be in a p orbital to 
complete the conjugation) contributes a fur- 
ther two making six altogether. The energy level 
diagram shows us that six 7C electrons complete- 
ly fill the bonding molecular orbitals thereby 
giving a stable structure. 

Aromatic heterocyclic compounds 

So far all the aromatic compounds you have seen have been hydrocarbons. However, most aromatic 
systems are heterocyclic — that is, involving atoms other than carbon and hydrogen. A simple exam- 
ple is pyridine. 

In this structure a nitrogen replaces one of the CH groups in benzene. The ring still has three 
double bonds and thus six 7C electrons. Consider the structure shown below, pyrrole. This is also aro- 
matic but how can we count six % electrons? 

In the cyclopentadiene ring above, there were also two double bonds and on deprotonation one 
carbon could formally contribute the other two electrons needed for aromaticity. In pyrrole the 
nitrogen's lone pair can make up the six Jt electrons needed for the system to be aromatic. 

We are really just beginning to scratch the surface of aromatic chemistry. You will meet 
many aromatic compounds in this book: in Chapter 22 we shall look at the chemistry of 
benzene and in Chapters 43 and 44 we shall discuss heterocyclic aromatic compounds. We shall 
finish off this chapter with a few more examples of some common aromatic compounds. In each 
case the aromatic part of the molecule — which may be one ring or several rings — is outlined in 
black. 




the anion has 671 electrons 
completely filling the bonding MOs 




o 



H 

pyrrole 



178 



7 ■ Derealization and conjugation 



First, a compound released by many cut plants, especially grasses, with a fresh delightful smell 
usually called 'new mown hay'. Coumarin is also present in some herbs such as lavender. It contains 
a benzene ring and an cc-pyrone fused together. 




°\^° 



coumarin - the smell 

of 'new mown hay' 

also found in lavender 





benzene 



a-pyrone 



Next, pirimicarb, a selective insecticide that kills sap-sucking aphid pests but does not affect the 
useful predators such as ladybirds (ladybugs) that eat them. It contains a pyrimidine ring — a ben- 
zene ring with two nitrogen atoms. 
Me 



Me 




pirimicarb - a selective 
insecticide which kills 
aphids but not ladybirds 




pyrimidine 



LSD stands for LySergic acid Diethylamide. It is the hallucinogenic drug 'acid'. When people walk 
off a building claiming that they can fly, they are probably on acid. It contains an indole ring made 
up of a benzene ring and a pyrrole ring fused together. 



Me 2 N 



NMe 
H 




LSD, lysergic acid diethylamide, 

the infamous 'acid' giving 

hallucinations and unfounded 

confidence in flying 




indole 



The world's best selling medicine in 1998 was Omeprazole, an antiulcer drug from Astra. It pre- 
vents excess acid in the stomach and allows the body to heal ulcers. It contains a pyridine ring and a 
benzimidazole ring, two aromatic heterocycles. 



OMe 



Me 




Omeprazole 

Astra's best selling 

antiulcer drug 



OMe 




benzimidazole 



The drug in the news in 1999 was Viagra, Pfizer's cure for male impotence. In the first three 
months after its release in 1998, 2.9 million prescriptions were issued for Viagra. It contains a 
simple benzene ring and a more complex heterocyclic system, which can be divided into two 
aromatic heterocyclic rings. 



Problems 



179 




Viagra 

Pfizer's treatment 

for male impotence 

(male erectile dysfunction) 




pyrimidone 



Me 

/ 
N 

> 



pyrazole 




benzene 



Finally, the iron compound haem, part of the haemoglobin molecule we use to carry oxygen 
around in our bloodstream. It contains the aromatic porphyrin ring system with its eighteen elec- 
trons arranged in annulene style. Chlorophyll, mentioned earlier in this chapter, has a similar aro- 
matic ring system. 




Me 



Me 



haem - part of the haemoglobin that 
transports oxygen in the blood 




porphyrin or porphin 

one 18 7t electron ring is shown in black. 

Others are possible 



Problems 



1. Are these molecules conjugated? Explain your answer in any 
reasonable way. 



K K K 

1 \/ 

K ^K ^K 

II 
K 


K 

1 
K 




1 II 
1 


1 1 
1 




Y* V K^ 

1 1 

1 


1 
K 


1 
K 




1 
K 



2. Draw a full orbital diagram for all the bonding and antibond- 
ing 71 orbitals in the three-membered cyclic cation shown here. 
The molecule is obviously very strained. Might it survive by also 
being aromatic? 
© 



A 



3. How extensive are the conjugated systems in these molecules? 




C0 2 H 

a p-lactam antibiotic 




OMe 



OMe 

the anti-cancer compound 
podophyllotoxin 

4. Draw diagrams to illustrate the conjugation present in these 
molecules. You should draw three types of diagram: (a) conjuga- 
tion arrows to give at least two different ways of representing the 
molecule joined by the correct 'reaction' arrow; (b) a diagram 
with dotted lines and partial charges (if any) to show the double 
bond and charge distribution (if any); and (c) a diagram of the 



180 



7 ■ Derealization and conjugation 



atomic orbitals that make up the lowest-energy bonding molecu- 
lar orbital. 



© 
NH 2 



H,N 






.0 




5. Which of these compounds are aromatic? Justify your answer 
with some electron counting. You are welcome to treat each 
ring separately or two or more rings together, whichever you 
prefer. 





© 

M 



H0 2 C, 



NH C0 2 H 




Me 



Me 



C0 2 H 




NHAc 



OMe 



methoxatin: co-enzyme from colchicine: compound from Autumn 
bacteria living on methane crocus used to treat gout 



C0 2 Me 




OH OH OH 

aklavinone: a tetracycline antibiotic 




6. A number of water-soluble pigments in the green/blue/violet 
ranges used as food dyes are based on cations of the type shown 
here. Explain why the general structure shows such long wave- 
length absorption and suggest why the extra functionality (OH 
group and sulfonate anions) is put into 'CI food green 4' a com- 
pound approved by the EU for use in food under E142. 



Me^ 



general structure for 
water soluble food 

dye in the 

green/blue/violet 

range 



Me^ 




green food dye 

'CI food green 4' 

[E142] 



7. Turn to Chapter 1 and look at the structures of the dyes in the 
shaving foam described on p. 000. Comment on the structures in 
comparison with those in Problem 6 and suggest where they get 
their colour from and why they too have extra functional groups. 
Then turn to the beginning of Chapter 1 (p. 000) and look at the 
structures of the compounds in the 'spectrum of molecules'. Can 
you see what kind of absorption leads to each colour? You will 
want to think about the conjugation in each molecule but you 
should not expect to correlate structures with wavelengths in any 
even roughly quantitative way. 

8. Go through the list of aromatic compounds at the end of the 
chapter and see how many electrons there are in the rings taken 
separately or taken together (if they are fused). Are all the num- 
bers of the (An + 2) kind? 



callistephin: natural red flower pigment 



Acidity, basicity, and p/C 



8 



Connections 


Building on: 




Arriving at: 


Looking forward to: 


• Conjugation and molecular stability 


• 


Why some molecules are acidic and 


• Acid and base catalysis in carbonyl 


ch7 




others basic 


reactions chl2 & chl4 


• Curly arrows represent derealization 


• 


Why some acids are strong and others 


• The role of catalysts in organic 


and mechanisms ch5 




weak 


mechanisms chl3 


• How orbitals overlap to form 


• 


Why some bases are strong and others 


• Making reactions selective using 


conjugated systems ch4 




weak 


acids and bases ch24 




• 


Estimating acidity and basicity using 
pH and pK a 






• 


Structure and equilibria in proton- 
transfer reactions 






• 


Which protons in more complex 
molecules are more acidic 






• 


Which lone pairs in more complex 
molecules are more basic 






• 


Quantitative acid/base ideas affecting 
reactions and solubility 






• 


Effects of quantitative acid/base 
ideas on medicine design 





Note from the authors to all readers 

This chapter contains physical data and mathematical material that some readers may find daunting. 
Organic chemistry students come from many different backgrounds since organic chemistry occu- 
pies a middle ground between the physical and the biological sciences. We hope that those from a 
more physical background will enjoy the material as it is. If you are one of those, you should work 
your way through the entire chapter. If you come from a more biological background, especially if 
you have done little maths at school, you may lose the essence of the chapter in a struggle to under- 
stand the equations. We have therefore picked out the more mathematical parts in boxes and you 
should abandon these parts if you find them too alien. We consider the general principles behind the 
chapter so important that we are not prepared to omit this essential material but you should try to 
grasp the principles without worrying too much about the equations. The ideas of acidity, basicity, 
and p_K" a values together with an approximate quantitative feel for the strength and weakness of acids 
and bases are at least as central for biochemistry as they are for organic chemistry. Please do not be 
discouraged but enjoy the challenge. 

Introduction 

This chapter is all about acidity, basicity, and piC a . Acids and bases are obviously important 
because many organic and biological reactions are catalysed by acids or bases, but what is pK a and 
what use is it? pK a tells us how acidic (or not) a given hydrogen atom in a compound is. This is useful 
because, if the first step in a reaction is the protonation or deprotonation of one of the reactants, it is 
obviously necessary to know where the compound would be protonated or deprotonated and what 
strength acid or base would be needed. It would be futile to use too weak a base to deprotonate a 
compound but, equally, using a very strong base where a weak one would do would be like trying to 



182 



8 ■ Acidity, basicity, and pK a 



crack open a walnut using a sledge hammer — you would succeed but your nut would be totally 
destroyed in the process. 

The aim of this chapter is to help you to understand why a given compound has the piC a that it 
does. Once you understand the trends involved, you should have a good feel for the pK a values of 
commonly encountered compounds and also be able to predict the values for unfamiliar com- 
pounds. 

Originally, a substance was identified as an acid if it exhibited the properties shown by other acids: 
a sour taste (the word acid is derived from the Latin acidus meaning 'sour') and the abilities to turn 
blue vegetable dyes red, to dissolve chalk with the evolution of gas, and to react with certain 'bases' to 
form salts. It seemed that all acids must therefore contain something in common and at the end of 
the eighteenth century, the French chemist Lavoisier erroneously proclaimed this common agent to 
be oxygen (indeed, he named oxygen from the Greek oxus 'acid' and gennao 'I produce'). Later it was 
realized that some acids, for example, hydrochloric acid, did not contain oxygen and soon hydrogen 
was identified as the key species. However, not all hydrogen-containing compounds are acidic, and 
at the end of the nineteenth century it was understood that such compounds are acidic only if they 
produce hydrogen ions H + in aqueous solution — the more acidic the compound, the more hydro- 
gen ions it produces. This was refined once more in 1923 by J.N. Brensted who proposed simple def- 
initions for acids and bases. 



Other definitions of acids and bases 
are useful, the most notable being 
those of Lewis, also proposed in 1923. 
However, for this chapter, the Br0nsted 
definition is entirely adequate. 



> Brensted definitions of acids and bases 

• An acid is a species having a tendency to lose a proton 

• A base is a species having a tendency to accept a proton 



-o 

i 



© 

'v'""H. 
H 



./ 



,0—v 



a structure for a solvated 

hydronium ion in water 

the dashed bonds represent 

hydrogen bonds 



Acidity 

An isolated proton is incredibly reactive — formation of H3O" 1 " in water 

Hydrochloric acid is a strong acid: the free energy AG° for its ionization equilibrium in water is -40 



kjmol 



HCI (aq) 



H + (aq) + Cl-(aq) AG° 



-40 kJ mol - 1 



Such a large negative AG° value means that the equilibrium lies well over to the right. In the gas 
phase, however, things are drastically different and AG° for the ionization is +1347 kJ moL . 

HCI (g) -^-— H + (g) + Cl-(g) AG° 298K = +1347 kJ mol-1 

This AG° value corresponds to 1 molecule of HCI in 10 being dissociated! This means that 
HCI does not spontaneously ionize in the gas phase — it does not lose protons at all. Why then 
is HCI such a strong acid in water? The key to this problem is, of course, the water. In the gas 
phase we would have to form an isolated proton (H + , hydrogen ion) and chloride ion and 
this is energetically very unfavourable. In contrast, in aqueous solution the proton is strongly 
attached to a water molecule to give the very stable hydronium ion, H 3 + , and the ions are no 
longer isolated but solvated. Even in the gas phase, adding an extra proton to neutral water is 
highly exothermic. 



H 2 0(g) +H + (g) 



H 3 + (g) AH" = - 686 kJ mor 1 



In fact, an isolated proton is so reactive that it will even add on to a molecule of methane in the gas 
phase to give CH5 in a strongly exothermic reaction (you have already encountered this species in 
mass spectrometry on p. 52). We are therefore extremely unlikely to have a naked proton in the gas 
phase and certainly never in solution. In aqueous solution a proton will be attached to a water mole- 
cule to give a hydronium ion, H 3 + (sometimes called a hydroxonium ion). This will be solvated 
just as any other cation (or anion) would be and hydrogen bonding gives rise to such exotic species as 
H9O4 (H 3 + -3H 2 0) shown here. 



Acidity 



183 



Every acid has a conjugate base 

In water, hydrogen chloride donates a proton to a water molecule to give a hydronium ion and chlo- 
ride ion, both of which are strongly solvated. 

HCI (aq) + H 2 (I) ^^ H 3 + (aq) + Cl _ (aq) 

In this reaction water is acting as a base, according to our definition above, by accepting a proton 
from HCI which in turn is acting as an acid by donating a proton. If we consider the reverse reaction 
(which is admittedly insignificant in this case since the equilibrium lies well over to the right), the 
chloride ion accepts a proton from the hydronium ion. Now the chloride is acting as a base and the 
hydronium ion as an acid. The chloride ion is called the conjugate base of hydrochloric acid and the 
hydronium ion, H30 + , is the conjugate acid of water. 



For any acid and any base 



AH + B 



BH + + A" 



where AH is an acid and A~ is its conjugate base and B is a base and BH + is its 
conjugate acid, that is, every acid has a conjugate base associated with it and every 
base has a conjugate acid associated with it. 

For example, with ammonia and acetic acid 

CH3COOH + NH 3 - ^-^ NH4 + CH3COO- 

the ammonium ion, NH4 , is the conjugate acid of the base ammonia, NH3, and the acetate ion, 
CH^COCF, is the conjugate base of acetic acid, CH3COOH. 

Water can behave as an acid or as a base 

If a strong acid is added to water, the water acts as a base and is protonated by the acid to become 
H3O . If we added a strong base to water, the base would deprotonate the water to give hydroxide 
ion, OH - , and here the water would be acting as an acid. Such compounds that can act as either an 
acid or a base are called amphoteric. 

With a strong enough acid, we can protonate almost anything and, 
likewise, with a strong enough base we can deprotonate almost anything. 
This means that, to a certain degree, all compounds are amphoteric. For 
example, hydrochloric acid will protonate acetic acid. 

In this example acetic acid is acting as a base! Other compounds need acids even stronger than 
HCI to protonate them. Remember that, in chemical ionization mass spectrometry (p. 52), proto- 
nated methane, CH5 , was used to protonate whatever sample we put in to the machine in order to 
give us a cation; CH5 is an incredibly strong acid. 

The amino acids you encountered in Chapter 2 are amphoteric. Unlike 
water, however, these compounds have separate acidic and basic groups 
built into the same molecule. 

When amino acids are dissolved in water, the acidic end protonates the 
basic end to give a species with both a positive and a negative charge on it. 
A neutral species that contains both a positive and a negative charge is 
called a zwitterion. an a min ° acid zwitterion 



M 



HCI 



OH 



© 
H,N 




.© 



© 



OH 



M 



CI 



© 



OH 



basic group 
H 2 N 




acidic group 



OH 



How the pH of a solution depends on the concentration of the acid 

You are probably already familiar with the pH scale: acidic solutions all pH 
have a pH of less than 7 — the lower the pH the more acidic the solution; 
alkaline solutions all have pHs greater than 7 — the higher the pH, the more 
basic the solution. Finally, pH 7 is neither acidic nor alkaline but neutral. 
The pH of a solution is only a measure of the acidity of the solution; it 



strongly 
acidic 



14 



weakly 
acidic 



neutral 



weakly 
basic 



strongly 
basic 



increasing acid strength increasing base strength 



184 



8 ■ Acidity, basicity, and pK a 



tells us nothing about how strong one acid might be relative to another. The pH of a solution of a 
given acid varies with its concentration: as we dilute the solution, the acidity falls and the pH increas- 
es. For example, as we decrease the concentration of HC1 in an aqueous solution from 1 to 0.1 to 0.01 
to 0.001 mol dm , the pH changes from to 1 to 2 to 3. 

What a pH meter actually measures is the concentration of hydronium ions in the solution. The 
scale is a logarithmic one and is defined as 

pH=-log[H 3 0+] 

Our solutions of HC1 above therefore have hydronium ion concentrations of [H30 + ] = 10 , 10~~ , 
10~ , and 10~ mol drrT respectively. Since the scale is logarithmic, a pH difference of 1 corresponds 
to a factor of 10 in hydronium ion concentration, a pH difference of 2 corresponds to a factor of 100, 
and so on. 

The ionization of water 

Pure water at 25 °C has a pH of 7.00. This means that the concentration of hydronium ions in water 
must be 10~~ mol dm~ (of course, it is actually the other way round: the hydronium ion concentra- 
tion in pure water is 10~~ mol dm~ ; hence its pH is 7.00). Hydronium ions in pure water can arise 
only from the self-dissociation or autoprotolysis of water. 
H 2 + H 2 -^-^ H 3 + (aq) + OH" (aq) 

In this reaction, one molecule of water is acting as a base, receiving a proton from the other, which 
in turn is acting as an acid by donating a proton. From the equation we see that, for every hydronium 
ion formed, we must also form a hydroxide ion and so in pure water the concentrations of hydroxide 
and hydronium ions are equal. 

[H 3 + ] = [OH - ] = 1CT 7 mol dm -3 

The product of these two concentrations is known as the ionization constant of water, _K\y (or as 
the ionic product of water, or maybe sometimes as the autoprotolysis constant, Kj^p) 

K w = [H 3 + ][OrT] = 1CT 14 mol 2 dm -6 at 25 °C 

This is a constant in aqueous solutions, albeit a very, very small one. This means that, if we know 
the hydronium ion concentration, we also know the hydroxide concentration and vice versa since 
the product of the two concentrations always equals 10~ . 

For example 

It is easy to work out the pH of a 0.1 M solution of sodium 
hydroxide, 

[Na0H] = 0.1M 

and, since the sodium hydroxide is fully ionized, 
[OH - ] = 0.1 M but [OH - ] x [H 3 + ] = 10~ 14 . 



Don't worry — water is still safe to drink 
despite all this acid and hydroxide in it! 
This is, of course, because the 
concentrations of hydronium and 
hydroxide ions are verysmall (10~ 7 
mol dm -3 corresponds to about 2 parts 
per billion). This very low concentration 
means that there are not enough free 
hydronium (or hydroxide) ions in water 
either to do us any harm when we drink 
it, or to catalyse chemical reactions. 



[H 3 + 


,_1°- 14 
0.1 


pH = - 


og[H 3 + ] 



10 13 mol dm 



-log(10" 



= 13 



How the pH of a solution also depends on the acid in question 

If we measured the pH of an aqueous solution of an organic acid and compared it to an equally concen- 
trated solution of HC1, we would probably find the pHs different. For example, whilst 0.1M HC1 has a 
pH of 1, the same concentration of acetic acid has a pH of 3.7 and is much less acidic. This can only 
mean that a 0.1M solution of acetic acid contains fewer hydronium ions than a 0.1M solution of HC1. 

# Aqueous hydrochloric acid (or any strong acid) has a lower pH than an equal 
concentration of aqueous acetic acid (or any weak acid) because it is more fully 
dissociated and thereby produces more hydronium ions. 

For hydrochloric acid, the equilibrium lies well over to the right: in effect, HC1 is completely dis- 
sociated. 

HCI (aq) + H 2 (I) =^= »- H 3 + (aq) + CP(aq) 



The definition of pK a 



185 



Acetic acid is not fully dissociated — the solution contains both acetic acid and acetate ions. 
CH3COOH (aq) + H 2 (I) . . H 3 + (aq) + CH 3 C00 _ (aq) 



Acids as preservatives 

Acetic acid is used as a preservative in many foods, for 
example, pickles, mayonnaise, bread, and fish products, 
because it prevents bacteria and fungi growing. However, 
its fungicidal nature is not due to any lowering of the pH of 
the foodstuff. In fact, it is the undissociated acid that acts 
as a bactericide and a fungicide in concentrations as low 
as 0.1-0.3%. Besides, such a low concentration has little 
effect on the pH of the foodstuff anyway. 



Although acetic acid can be added directly to a foodstuff 
(disguised as E260), it is more common to add vinegar 
which contains between 10 and 15% acetic acid. This 
makes the product more 'natural' since it avoids the nasty 
'E numbers'. Actually, vinegar has also replaced other 
acids used as preservatives, such as propionic 
(propanoic) acid (E280) and its salts (E281, E282, and 
E283). 



The definition of pK a 



Now we need to be clearer about 'strong' and 'weak' acids. In order to measure the strength of an 
acid relative to water and find out how effective a proton donor it is, we must look at the equilibrium 
constant for the reaction 

AH (aq) + H 2 (I) ^=== H 3 + (aq) + A" (aq) 

The position of equilibrium is measured by the equilibrium constant for this reaction K e „. 
K _ [H 3 + ][A-] 
6q [AH][H 2 0] 
The concentration of water remains essentially constant (at 55.56 mol dnT ) with dilute solutions 
of acids wherever the equilibrium may be and a new equilibrium constant, K a , is defined and called 
the acidity constant. 

[H 3 + ][A"] 



K a 



[AH] 



Like pH, this is also expressed in a logarithmic form, piC a . 

• pK a = -logK a 

Because of the minus sign in this definition, the lower the pK a , the larger the 
equilibrium constant, K a , is and hence the stronger the acid. ThepK a of the acid is 
thepH where it is exactly half dissociated. At pHs above the pK a , the acid HA exists 
as A~ in water; at pHs below the pK a , it exists as undissociated HA. 

At pHs above the pK a of the acid, it will also be more soluble in water. Hydrocarbons are insoluble 
in water — oil floats on water, for example. Unless a compound has some hydrophilic groups in it 
that can hydrogen bond to the water, it too will be insoluble. Ionic groups considerably increase a 
compound's solubility and so the ion A - is much more soluble in water than the undissociated acid 
HA. In fact water can solvate both cations and anions, unlike some of the solvents you will meet later. 
This means that we can increase the solubility of a neutral acid in water by increasing the proportion 
of its conjugate base present. All we need to do is raise the pH. 

A simple example is aspirin: whilst the acid itself is not very soluble in water, the sodium salt is 
much more soluble (soluble aspirin is actually the sodium or calcium salt of 'normal' aspirin). 

Conversely, if the pH of a solution is 



HO^O 



© 
Na 




0> 



Me 



Me 



aspirin 
not very soluble in water 



(XV 



the sodium (or calcium) salt of 
aspirin is more soluble in water 



lowered, the amount of the acidic form 
present increases, and the solubility 
decreases. In the acidic environment of 
the stomach (around pH 1-2), soluble 
aspirin will be converted back to the 
normal acidic form and precipitate out 
of solution. 



How concentrated is water ? Very 
concentrated you may say — but 
the concentration is limited. We 
know that one mole of pure water 
has a mass of 18 g and occupies 
18 cm 3 . So, in one dm 3 , there are 
1000/18 = 55.56 mol. You 
cannot get more concentrated 
water than this (unless you did 
something drastic like taking it 
into a black hole!). 



This is how we can work out that 
the pK a of the acid is the pH at 
which it is exactly half 
dissociated: we can rearrange the 
equation for K a to give 

[AH] 



[H 3 + ] = K a 



[A" 



Taking minus the log of both sides 
gives us 



pH = pK a + log 



w 



J AH l 

If the concentrations of acid AH 
and its conjugate base A~ are 
equal, the term in brackets 
equals 1 and log (1) = and so 
the pH simply equals the pK a of 
the acid. 

This means that, if we took a 0.1 
M aqueous solution of acetic acid 
and raised its pH from 3.7 (its 
natural pH) to 4.76 (the pK a of 
acetic acid) using dilute sodium 
hydroxide solution, the resultant 
solution would contain equal 
concentrations of acetic acid and 
acetate ion. 



186 



8 ■ Acidity, basicity, and pK a 



In the same way, organic bases such as amines can be dissolved by loweringihe pH. Codeine (7,8- 
didehydro-4,5-epoxy-3-methoxy-17-methylmorphinan-6-ol) is a commonly used painkiller. 
Codeine itself is not very soluble in water but it does contain a basic nitrogen atom that can be pro to - 
nated to give a more soluble salt. It is usually encountered as a phosphate salt. The structure is com- 
plex, but that doesn't matter. „ 

\© 
N— Me 




neutral codeine 
sparingly soluble in water 



the conjugate acid is much 
more soluble in water 



Charged compounds can be separated by acid-base extraction 

Adjusting the pH of a solution often provides an easy way to separate compounds. Since weak acids 
form soluble anions at pHs above their p_K" a values, this presents us with an easy method for extract- 
ing organic acids from mixtures of other compounds. For example, if we dissolve the mixture of 
compounds in dichloromethane (which is immiscible with water, that is, it will not mix with water 
but instead forms a separate layer) and 'wash' this solution with aqueous sodium hydroxide, any 
organic acids present will be converted to their water-soluble salts and dissolve into the water layer. 
We have extracted the organic acids into the aqueous layer. If we then separate and acidify the aque- 
ous layer, the acid form, being less soluble in water, will precipitate out. If the acid form has a charge 
and the conjugate base is neutral as with amines, for example, now the cationic acid form will be 
more soluble in water than the conjugate base. 

• Acid-base extraction 

For a neutral weak organic acid HA 

HA(aq) + H 2 ^^ H 3 0®(aq) + A e (aq) 

• Anionic A~ is more soluble in water than the neutral acid HA 

• Neutral acid HA is more soluble in organic solvents than anionic A~ 
For a neutral weak organic base B 



HB"(aq) + H 2 



H 3 w (aq) + B(aq) 



• The cationic acid HB + is more soluble in water than the neutral conjugate base B 

• The neutral conjugate base, B is more soluble in organic solvents than the 
cationic acid HB + 

Separating a mixture of benzoic acid (PI1CO2H) and toluene (PhMe) is easy: dissolve the mixture 
in CH2CI2, add aqueous NaOH, shake the mixture of solutions, and separate the layers. The CH2CI2 
layer contains all the toluene. The aqueous layer contains the sodium salt of benzoic acid. Addition 
of HC1 to the aqueous layer precipitates the insoluble benzoic acid. 




Me 




C0 2 H 



NaOH 




Me 




C0° 



Na 



© 



insoluble 
in water 



insoluble 

in water 



insoluble 
in water 



soluble 
in water 



In the same way, any basic compounds dissolved in an organic layer could be extracted by wash- 
ing the layer with dilute aqueous acid and recovered by raising the pH, which will precipitate out the 
less soluble neutral compound. 



The definition of pK a 



187 



Whenever you do any extractions or washes in practical experiments, just stop and ask yourself: 
'What is happening here? In which layer is my compound and why?' That way you will be less likely 
to throw the wrong layer (and your precious compound) away! 



Benzoic acid preserves soft drinks 

Benzoic acid is used as a preservative in foods and soft 
drinks (E210). Like acetic acid, it is only the acid form that 
is effective as a bactericide. Consequently, benzoic acid 
can be used as a preservative only in foodstuffs with a 
relatively low pH, ideally less than its pK a of 4. 2. This isn't 
usually a problem: soft drinks, for example, typically have 



a pH of 2-3. Benzoic acid is often added as the sodium 
salt (E211), perhaps because this can be added to the 
recipe as a concentrated solution in water. At the lowpH in 
the final drink, most of the salt will be protonated to give 
benzoic acid proper, which presumably remains in 
solution because it is so dilute. 



A graphical description of the pK a of acids and bases 

For both cases, adjusting the pH alters the proportions of the acid form and of the conjugate base. 
The graph plots the concentration of the free acid AH (green curve) and the ionized conjugate base 
A~ (red curve) as percentages of the total concentration as the pH is varied. 



i 


li 






i 


100 % 




-----^ 




< 




< 


\ 




CD 
U) 

co 




-o 
'o 

(0 


V 




CD 
CO 
■Qfl 

'c 1 


50% 


CD 


mainly AH 


mainly A 




4-1 

CD 
O 

CD 

Q. 


/ \ 


\ 


o 
o 

CD 

■qd 

CO 

c 

CD 
O 

CD 


0% 




-""'' [HA] : 


= [A"] **^ mm 





100 % 



E 50 % 



0% 



low pH 



pH = pK a 



high pH 



At low pH the compound exists entirely as AH and at high pH entirely as A - . At the ipK a the con- 
centration of each species, AH and A - , is the same. At pHs near the pK a the compound exists as a 
mixture of the two forms. 

An acid's pK a depends on the stability of its conjugate base 

HC1 is a much stronger acid than acetic acid: the pK a of HC1 is around -7 compared to 4.76 for acetic 
acid. This tells us that in solution K a for hydrogen chloride is 10 moldm whilst for acetic acid it is 
only 10 = 1.74 x 10 moldm . Why are the equilibria so different? Why does hydrogen chlo- 

ride fully dissociate but acetic acid do so only partially? 

HCI (aq) + H 2 (I) ^ w h 3 + (aq) + Cr(aq) K a = 10 7 

CH3COOH (aq) + H 2 (I) ^ H 3 + (aq) + CH 3 COO _ (aq) K a = 1.74 x 10 5 

The answer must have something to do with the conjugate base A~ of each acid HA, since this is 
the only thing that varies from one acid to another. In both the equilibria above, water acts as a base 
by accepting a proton from the acid. For the hydrochloric acid equilibrium in the reverse direction, 
the chloride ion is not a strong enough base to deprotonate the hydronium ion. Acetate, on the other 
hand, is easily protonated by H3O to give neutral acetic acid, which means that acetate must be a 
stronger base than chloride ion. 

• Acid and conjugate base strength 

• The stronger the acid HA, the weaker its conjugate base, A~ 

• The stronger the base A~, the weaker its conjugate acid AH 



188 



8 ■ Acidity, basicity, and pK a 



For example, hydrogen iodide has a very low pK a of-10. This means that HI is a strong enough 
acid to protonate most things. Its conjugate base, iodide ion, is therefore not very basic at all — in 
fact, we very rarely think of it as a base — it will not deprotonate anything. A very powerful base is 
methyllithium, MeLi. Here we effectively have CH3 (but see Chapter 9), which can accept a proton 
to become neutral methane, CH4. Methane is therefore the conjugate acid in this case. Clearly, 
methane isn't at all acidic — its p_K" a is about 48. 

Table 8.1 gives a list of compounds and their 
approximate pK a values. 

Over the next few pages we shall be considering 
the reasons for these differences in acid strength but 
we are first going to consider the simple conse- 
quences of mixing acids or bases of different strength. 



An alternative way of looking at this 
is that chloride ion is much happier 
being a chloride ion than acetate is 
being an acetate ion: the chloride 
ion is fundamentally more stable 
than is the acetate ion. 



Have a close look at Table 8.1 for 
there are some interesting points 
to notice. 

• Look at the acids themselves — 
we have neutral, cationic, and 
even anionic acids 

• Notice the range of different 
elements carrying the negative 
charge of the conjugate 
bases — we have iodine, 
chlorine, oxygen, sulfur, 
nitrogen, and carbon and many 
more are possible 

• Most importantly, notice the 
vast range of pK a values: from 
around -10 to 50. This 
corresponds to a difference of 
10 60 in the equilibrium 
constants and these are by no 
means the limits. Other 
compounds or intermediates 
can have pK a values even 
greater or less than these. 



That the difference in pK a s gives the 
log of the equilibrium constant can 
easily be shown by considering, as an 
example, the equilibrium for the 
reaction between hydrogen sulfate and 
acetate. 

HSO4 + CH 3 00 _ (aq) ^ CH 3 C00H(aq) - 



The difference in pi<C a values tells us the 
equilibrium constant between two acids or 
bases 

If we have a mixture of two bases in a pot and we throw 
in a few protons, where will the protons end up? 
Clearly, this depends on the relative strengths of the 
bases — if they are equally strong, then the protons will 
be shared between them equally. If one base is stronger 
than the other, then this base will get more than its fair 
share of protons. If we put into our pot not two bases 
but one base and an acid, then it's exactly the same as 
putting in two bases and then adding some protons — 
the protons end up on the strongest base. Exactly how 
the protons are shared depends on the difference in 
strengths of the two bases, which is related to the dif- 
ference in the pK a s of their conjugate acids. 



Table 8.1 The pK a value 


of some compound 


Acid 


P«a 


Conjugate base 


HI 


ca. -10 


r 


HCI 


ca.-7 


cr 


H 2 S0 4 


ca.-3 


HSO4 


HSO4 


2.0 


S0|" 


CH3COOH 


4.8 


CH3C00- 


H 2 S 


7.0 


HS" 


NHj 


9.2 


NH 3 


C 6 H 5 0H 


10.0 


c 6 H 5 cr 


CH3OH 


15.5 


CH30- 







00 



CH 



r^cHi 



20.0 



CH 



i CH 2 



NH 3 

CeHg 

CH 4 



25 


CH= 


33 


NH2 


ca. 43 


C 6HJ5 


ca.48 


CHI 



c © 



# In a mixture of two acids 


or two bases 






• The ratio of K a values 
the reaction between a 


gives us an indication of the 
base and an acid 


equilibrium constant for 


• The difference in piC a s 


gives us the 


log of the equilibrium constant. 



-|2- 



•eq 



[S0^"][CH 3 C00H] 
[HSO4HCH3COO"] 



The equilibrium constant for this reaction is simply the K a for the 
hydrogen sulfate equilibrium divided by the K a for the acetic acid 
equilibrium. 



K 



eq 



[SOf HCH3COOH] 
[HSO4HCH3COO"] 



:K a (HS04)x- 



[H 3 + ][S02- 



[HSO4 
10 



[CH3COOH] 
[H 3 + ][CH 3 C00~ 



-2 



i-4.« 



= 10^ 



= 600 



As an example, let us look at a method for acetylating aromatic amines in 
aqueous solution. This reaction has a special name — the Lumiere-Barbier 
method. We shall consider the acetylation of aniline PI1NH2 (a basic aromat- 
ic amine) using acetic anhydride. The procedure for this reaction is as follows. 

1 Dissolve one equivalent of aniline in water to which one equivalent of 
hydrochloric acid has been added. 

Aniline is not soluble in water to any significant degree. This isn't surprising 
as aniline is just a hydrophobic hydrocarbon with an amine group. The HCI 
(p-Ka - 7) protonates the aniline (pK a of the conjugate acid of aniline is 4.6) to 
give the hydrochloride. Now we have a salt that is very soluble in water. 



K a (CH 3 C00H) 10 " 

This tells us in our case that, if we mixed sodium hydrogen sulfate and 
sodium acetate in water, we would end up with mainly sodium sulfate 
and acetic acid, the equilibrium constant for the reaction above being 
approximately 600. 



.C\ 



NH, 



aniline 
insoluble in water 




H J -0 



0H 2 

© 




© 
NH 3 



OH, 



anilinium ion 
soluble in water 



The definition of pK a 



189 



2 Warm to 50°C and add 1.2 equivalents of acetic anhydride followed by 1.2 equivalents of aqueous 
sodium acetate solution. 

The acetic anhydride could be attacked by either the water, the acetate, or by aniline itself. Aniline 
is much more nucleophilic than the other two nucleophiles but only aniline itself can attack the 
anhydride: protonated aniline has no lone pair and is not nucleophilic. This, then, is the role of the 
sodium acetate — to act as a base and deprotonate the aniline hydrochloride. The pi^s of the aniline 
hydrochloride and acetic acid are about the same, around 4.7. An equilibrium will be set up to give 
some neutral aniline which will then attack the acetic anhydride and form the amide. 

© 
BH 




O* 

aniline attacks it* of acetic anhydride 




deprotonation of ammonium ion 






acetate is a good leaving group 



the product - an amide 



3 Cool in ice and filter off crystals of product, acetanilide. 

The product is insoluble in water and, because it is an amide, is much less basic than aniline (pK a 
of conjugate acid < 0) and so is not protonated to give a water-soluble salt. 



More from pK a s: Calculating the pK a values for water acting as a base and as an 
acid 

The material in this box is quite mathematical and maybe skipped if you find it too alien. 



How easy is it to protonate or deprotonate water? 

All our reactions so far have been in water and it is easy to 
forget that water itself also competes for protons. If, for 
example, we have both sulfuric acid H2SO4 and 
hydrochloric acid HCI in aqueous solution, hydrochloric 
acid with its lower pK a (-7) will not protonate the conjugate 
base of sulfuric acid (pK a -3), hydrogen sulfate HSO4: both 
acids will protonate water instead. So water is a stronger 
base than either chloride or hydrogen sulfate ions. In fact, 
we can work out the pK a forthe protonation of water. 

We want to answer the questions: 'How easy is it to 
protonate water? What strength of acid do we need?' 

Look at this simple reaction: 

H 3 + (aq) + H 2 0(l) ^=^ H 2 0(l) + H 3 + (aq) 

Obviously, the equilibrium constant for the equation above 
will be 1 since both sides of the equation are the same. 
But we don't use normal equilibrium constants — we use 
the acidity constant, K a , which is slightly different. 
Remember that this is actually the normal equilibrium 
constant for the reaction multiplied by [H2O] = 55.56, the 
'concentration' of water. This is normally useful in that it 
cancels out the [H20]term in the denominator but not in 
this case. 

Here we have 



:K eq x[H 2 0]: 



[H 2 0][H 3 + ] 



x [H 2 0] = [H 2 0] = 55.56 



[H 2 0][H 3 + ] 

so the pK a forthe protonation ofwateris: pK a (H30 + ) = 
-log(55.56) = -1.74. 

The pK a also equals the pH when we have equal 



concentrations of acid and conjugate base. A solution 
would be quite acidic if exactly half of the number of water 
molecules present were hydronium ions — its pH would be 
-1.74. 

So with water acting as a base 

AH(aq) + H 2 0(l) ^=^ H 3 + (aq) + A(aq) 

It is clear that for any acid with a lower pK a than -1.74, the 
equilibrium will lie over to the right. 

• Acids with a lower pKa than -1.74 will protonate 
water completely 

We can also work out the pK a for water acting as an acid. 
Now the equilibrium is 

H 2 0(l) + H 2 0(l) ^=^ H 3 + (aq) + OH(aq) 

Going through the same calculations as before, we find 
[H 3 + ][0H" 



K a = K eq x[H 2 0] = 



10" 



[H 2 0][H 2 0] 



- x [H 2 0] 



[H 2 0] 55.56 



= 1.80x10" 



so the pK a forthe deprotonation of water is: pK a (H 2 0) = 
- Iog(1.80 x 10~ 16 ) = 15.74. 

This means that, if we put in water a base whose 
conjugate acid's pK a is greater than 15.74, it will simply 
be protonated by the water and give an equivalent amount 
of hydroxide ions. 

• Bases B whose conjugate acid HB has a higher pK a 
than 15.74 will deprotonate water completely 



Any sharp-eyed readers may notice an 
inconsistency in the statement that the 
pK a equals the pH when we have equal 
concentrations of acid and conjugate 
base. If, when [A - ] = [AH], the pK a = pH, 
then the pK a for water equals the pH 
when [H 2 0] = [H 3 + j. We assume that 
[H 2 0] is constant at 55.56 mol dm" 3 
and so [H 3 + ] must also equal 55.56 
mol dm -3 and hence pH = pK a = 
-log(55.56). This assumption cannot 
be valid here: rather [H 2 0] + [H 3 + ] 
should equal approximately 55.56 
mol dm -3 . 



190 



8 ■ Acidity, basicity, and pK a 



k The strongest base in aqueous solution is OH and the strongest acid in aqueous 
solution is H 3 + . Remember that: 

• Addition of stronger bases than OH~ just gives more OH~ by the deprotona- 
tion of water 

• Addition of stronger acids than H30 + just gives more H30 + by protonation 
of water 

Also remember that: 

• The pH of pure water at 25°C is 7.00 (not the pK a ) 

• The pK a of H 2 is 15.74 

• The pK a of H 3 0+ is -1.74 



.© 



H 3 



H 2 



HoO 



HO 



,© 



pH <-1.74 pH -1.74 pH 7 

strongly acidic neutral 



pH 15.74 pH >15.74 

strongly basic 



Because the pK a values for very 
strong acids and bases are so 
hard to determine, you will find 
that they often differ in different 
texts — sometimes the values are 
no better than good guesses! 
However, while the absolute 
values may differ, the relative 
values (which is the important 
thing because we need only a 
rough guide) are usually 
consistent. 



The choice of solvent limits the pK a range we can use 

In water, our effective pK a range is only -1.74 to 15.74, that is, it is determined by the solvent. This is 
known as the levelling effect of the solvent. This is an important point. It means that, if we want to 
remove the proton from something with a high pK a , say 25-30, it would be impossible to do this in 
water since the strongest base we can use is hydroxide. If we do need a stronger base than OH~, we 
must use a different solvent system. 

For example, if we wanted to deprotonate ethyne (acetylene, pK a 25), then hydroxide (the 
strongest base we could have in aqueous solution, pK a 15.7) would establish an equilibrium where 



only linlO' (10 '/10 ) ethyne molecules were deprotonated. This means about 1 in 2 billion of 

our ethyne molecules will be deprotonated at any one time. Since, no matter what base we dissolve 

in water, we will only at best get hydroxide ions, this is the best we could do in water. So, in order 

to deprotonate ethyne to any appreciable extent, we must use a different solvent that does not have a 

pK a less than 25. Conditions often used to do this reaction are sodium amide (NaNTy in liquid 

ammonia. 

NH 3 (I) 

CI^=C — H + NH® ^= *"- CH=C8 + NH 3 



Using the pK a s of NH 3 (ca. 33) and ethyne (25) we would predict an equilibrium constant for this 
reaction of 10 (10 _ /10~ ) — well over to the right. Amide ions can be used to deprotonate 
alkynes. 

Since we have an upper and a lower limit on the strength of an acid or base that we can use 
in water, this poses a bit of a problem: How do we know that the pK a for HC1 is greater than that 
of H2SO4 if both completely protonate the water? How do we know that the pK a of methane 
is greater than that of ethyne since both the conjugate bases fully deprotonate water? The answer is 
that we can't simply measure the equilibrium for the reaction in water — we can do this only for 
piC a s that fall between the pK a values of water itself. Outside this range, p_K" a values are determined 
in other solvents and the results are extrapolated to give a value for what the pK a in water might 
be. 

Constructing a pK a scale 

We now want to look at ways to rationalize the different pK a values for different compounds — we 
wouldn't want to have to memorize all the values. You will need to get a feel for the p-K" a values of 



The definition of pK a 



191 



different compounds and, if you know what factors affect them, it will make it much easier to predict 
an approximate pK a value, or at least understand why a given compound has the pK a value that it 
does. 

A number of factors affect the strength of an acid, AH. 

AH (solvent) .. — A" (solvent) + H + (solvent) 

These include: 

1 Intrinsic stability of the conjugate base, anion A - . Stability can arise, for example, by having 
the negative charge on an electronegative atom or by spreading the charge over other groups. 
Either way, the more 'stable' the conjugate base, the less basic it will be and so the stronger the 
acid 

2 Bond strength A-H. Clearly, the easier it is to break this bond, the stronger the acid 

3 The solvent. The better the solvent is at stabilizing the ions formed, the easier it is for the reaction 
to occur 

• Acid strength 

• The most important factor in the strength of an acid is the stability of the 
conjugate base — the more stable the conjugate base, the stronger the 
acid 

• An important factor in the stability of the conjugate base is which 
element the negative charge is on — the more electronegative the element, 
the more stable the conjugate base 



The negative charge on an electronegative element stabilizes the conjugate base 

The pi<T a values for second row hydrides CH4, NH3, H2O, and HF are about 48, 33, 16, and 3, respec- 
tively. This trend is due to the increasing electronegativities across the period: F~ is much more stable 
than CH3, because fluorine is much more electronegative than carbon. 

Weak A-H bonds make stronger acids 

However, on descending group VII (group 17), the p_K" a values for HF, HC1, HBr, and HI decrease in 
the order 3, -7, -9, and -10. Since the electronegativities decrease on descending the group we might 
expect an increase in P-K" a s. The decrease observed is actually due to the weakening bond strengths on 
descending the group and to some extent the way in which the charge can be spread over the increas- 
ingly large anions. 

Derealization of the negative charge stabilizes the conjugate base 

The acids HCIO, HCIO2, HCIO3, and HCIO4 have pK a values 7.5, 2, — 1, and about -10, respectively. 
In each case the acidic proton is on an oxygen attached to chlorine, that is, we are removing a proton 
from the same environment in each case. Why then is perchloric acid, HCIO4, some 17 orders of mag- 
nitude stronger in acidity than hypochlorous acid, HCIO? Once the proton is removed, we end up 
with a negative charge on oxygen. For hypochlorous acid, this is localized on the one oxygen. With 
each successive oxygen, the charge can be more delocalized, and this makes the anion more 
stable. For example, with perchloric acid, the negative charge can be delocalized over all four oxygen 
atoms. 



rO 



I 

0# /N 



© 



BH 



0©^ 

IV 



etc. 



o*fq — o^Kfe 



0/ 




the negative charge on the perchlorate anion can be delocalized equally over all four oxygens 



That the charge is spread out over 
all the oxygen atoms equally is 
shown by electron diffraction 
studies: whereas perchloric acid 
has two types of CI-0 bond, one 
163.5 pm and the other three 
140.8 pm long, in the perchlorate 
anion all CI-0 bond lengths are 
the same, 144 pm, and all 
O-CI-0 bond angles are 109.5°. 



192 



8 ■ Acidity, basicity, and pK a 



Similar arguments explain the p-K" a s for other oxygen acids, for example, ethanol (p-fC a , 15.9), 
acetic acid (4.8), and methane sulfonic acid (-1.9). In ethoxide, the negative charge is localized on 
one oxygen atom, whilst in acetate the charge is delocalized over two oxygens and in methane sul- 
fonate it is spread over three oxygens. 

ethoxide acetate 



Just to remind you: these derealization 
arrows do not indicate that the charge 
is actually moving from atom to atom. 
These structures simply show that the 
charge is spread out in the molecular 
orbitals and mainly concentrated on the 
oxygen atoms. 



K H 



M 








charge localized on 
one oxygen 




charge delocalized over two oxygens 




Ho 





i 



methane sulfonate 



Mcr"" ^0 



,J 




charge delocalized over three oxygens 



In phenol, PhOH, the OH group is directly attached to a benzene ring. On deprotonation, the 
negative charge can again be delocalized, not on to other oxygens but this time on to the aromatic 
ring itself. 

'" Q© p /H 0© 0©) p 




cyclohexanol 
pK a 16 







phenol 
pK a 10 



phenoxide 



derealization increases the 
electron density on the ring 



the lone pair in the p orbital can overlap with the it system of the ring, 
spreading the negative charge on the oxygen on to the benxene ring 




Q 



these two lone pairs are in sp 2 orbitals and 
do not overlap with the % system of the ring 



The effect of this is to stabilize the phenoxide anion relative to the conjugate base of cyclohexanol 
where no derealization is possible and this is reflected in the P-KaS of the two compounds: 10 for 
phenol but 16 for cyclohexanol. 



• Get a feel for pK 


a s! 








Notice that these oxygen acids have pK a s that 


conveniently fall in 


units of 5 


(approximately). 










Acid 


RS0 2 OH 


RC0 2 H 


ArOH 


ROH 


Approx. pK a 





5 


10 


15 



The same derealization of charge can stabilize anions derived from deprotonating carbon acids. 
These are acids where the proton is removed from carbon rather than oxygen and, in general, they 
are weaker than oxygen acids because carbon is less electronegative. If the negative charge can be 
delocalized on to more electronegative atoms such as oxygen or nitrogen, the conjugate base will be 
stabilized and hence the acid will be stronger. 

Table 8.2 shows a selection of carbon acids with their conjugate bases and p-JC a s. In each case the 
proton removed is shown in black. 



The definition of pK a 



193 



Table 8.2 The conjugate bases and pK a s of some carbon acids 
Acid Conjugate base pK a Comments 



H h 




H H 




CH 



X, 



CH, 



CH/' © ^2 



-50 charge is localized on one carbon — difficult since carbon is 
not very electronegative 



-43 charge is delocalized over jc system — better but still not really 
good 



13.5 charge is delocalized over jt system but is mainly on the 
electronegative oxygen — much better 




charge delocalized over % system but mainly over two 
oxygens — better still 



*C0 



-48 charge is localized on one carbon — again very unsatisfactory 






H 2 C=N 



10 



charge is delocalized but mainly on oxygens of nitro group 



o 2 N 



^ 



NO, 



NO, 



charge can be delocalized overtwo nitro groups — more stable 
anion 



H 
2 N 



?- 



N0 2 



2 N 



N0 2 



charge can be delocalized over three nitro groups — very 
stable anion 



It isn't necessary for a group to be conjugated in order to spread the negative charge: any group 
that withdraws electrons will help to stabilize the conjugate base and therefore increase the strength 
of the acid. Some examples are shown below for both oxygen and carbon acids. 

electron-withdrawing groups lowering the pK a of carboxylic acids 
0,0 



0,N 



© 
Me 3 N 



"OH ~OH ^OH 

pK a 4.76 pK a 1.7 pK a 1.8 

electron-withdrawing groups lowering the pK a of alcohols 



N 



CF, 




OH 

pK a 2.4 



OH 

pK a 3.6 



H 3 C — OH 

pK a 15.5 



F 3 C^ ^OH 

pK a 12.4 



F 3 C- XF 3 



F 3 C^ ^OH 

pK a 9.3 



F 3 C^ ^OH 

pK a 5.4 



Picric acid is a very acidic phenol 

Electron-withdrawing effects on aromatic rings 
will be covered in more detail in Chapter 22 but 
for the time being note that electron-withdrawing 
groups can considerably lower the pK a s of 
substituted phenols and carboxylic acids, as 
illustrated by picric acid. 



0,N 



2, 4, 6-Trinitrophenol's more common name, 
picric acid, reflects the strong acidity of this 
compound (pK a 0.7 compared to phenol's 10.0). 
Picric acid used to be used in the dyeing industry 
but is little used now because it is also a powerful 
explosive (compare its structure with that of TNT!). 




• Notice how very electron- 
withdrawing the nitro group is — 
it lowers the pK a of acetic acid 
even more than a quaternary 
ammonium salt! 

• Notice also that the fourth 
alcohol with three CF 3 groups is 
almost as acidic as acetic acid. 

electron-withdrawing groups lowering 
the pK a s of carbon acids 

F 3 C, H F 3 C, CF 3 

V %/ FaC— H 

F 3 C H F 3 C H fiuoroform 

pK a ca. 22 pK a ca. 10 pK a 26 



0,N 



N0 2 

trinitrotoluene, TNT 




194 



8 ■ Acidity, basicity, and pK a 



Electron withdrawal in these molecules is the result of bond polarization from an inductive 
effect (Chapter 5). The electrons in a bond between carbon and a more electronegative element 
such as N, O, or F will be unevenly distributed with a greater electron density towards the more elec- 
tronegative atom. This polarization is passed on more and more weakly throughout the carbon 
skeleton. The three fluorine atoms in CF3H reduce the pK a to 26 from the 48 of methane, while the 
nine fluorines in (CF3)3CH reduce the p_K" a still further to 10. 

Such inductive effects become less significant as the electron-withdrawing group gets further away 
from the negative charge as is shown by the pX" a s for these chlorobutanoic acids: 2-chloro acid is signif- 
icantly stronger than butanoic acid but by the time the chlorine atom is on C4, there is almost no effect. 
CI 



Study carefully the pK a sforthe 
haloform series, CHX 3 — they may 
not do what you think they should! 
Chloroform is much more acidic 
than fluoroform even though 
fluorine is more electronegative 
(likewise with bromoform and 
chloroform). The anion CF3 must 
be slightly destabilized because of 
some backdonation of electrons. 
The anion from chloroform and 
bromoform may also be stabilized 
by some interaction with the d 
orbitals (there aren't any on 
fluorine). The conjugate base 
anion of bromoform is relatively 
stable — you will meet this again in 
the bromoform/iodoform reaction 
(Chapter 21). 



H H 

pK a ca. 50 

■ H mo 



H 




H 



lone pair of CH 3 CH 2 
in sp 3 orbital 







1 ^OH 




OH 



OH 



pK a 4.8 



CI 

pK a 2.8 




OH 



CI 



pK a 4.1 



pK a 4.5 



H 



H 



Hybridization can also affect the pK a 

The hybridization of the orbital from which the proton is removed also affects the p_K" a . Since s 
orbitals are held closer to the nucleus than are p orbitals, the electrons in them are lower in energy, 
that is, more stable. Consequently, the more s character an orbital has, the more tightly held are the 

electrons in it. This means that electrons in an sp 
orbital (50% s character) are lower in energy than 
those in an sp orbital (33% s character), which are, in 
turn, lower in energy than those in an sp orbital (25% 
s character). Hence the anions derived from ethane, 
ethene, and ethyne increase in stability in this order 
and this is reflected in their p-K" a s. Cyanide ion, ~CN, 
with an electronegative element as well as an sp 
hybridized anion, is even more stable and HCN has a 
piC a of about 10. 



X 

H H 

pK a ca. 44 



pK a ca. 26 



H x. 







o © 



© 



lone pair of CH 2 =CH 
in sp 2 orbital 



lone pair of HC^=C 
in sp orbital 







More remote hybridization is 
also important 

The more s character an orbital has, the more 
it holds on to the electrons in it. This makes 
an sp hybridized carbon less electron- 
donating than an sp 2 one, which in turn is 
less electron-donating than an sp 3 carbon. 
This is reflected in the pK a s of the compounds 
shown here. 




v OH 



pK a 4.9 



pK a 4.2 



pK a 4.2 



pK a 13.5 



~< 



pK a 1.9 



Highly conjugated carbon acids 

If we can delocalize the negative 
charge of a conjugate anion on to 
oxygen, the anion is more stable and 
consequently the acid is stronger. 
Even derealization on to carbon 
alone is good if there is enough of it, 
which is why some highly delocal- 
ized hydrocarbons have remarkably 
low piC a s for hydrocarbons. Look at 
this series . 



H 




pK a ca. 48 pK a ca. 40 



pK a ca. 33 



pK a ca. 32 



The definition of pK a 



195 



Increasing the number of phenyl groups decreases the pK a — this is what we expect, since we can 
delocalize the charge over all the rings. Notice, however, that each successive phenyl ring has less 
effect on the pi<f a : the first ring lowers the p_K" a by 8 units, the second by 7, and the third by only 1 unit. 
In order to have effective derealization, the system must be planar (Chapter 7). Three phenyl rings 
cannot arrange themselves in a plane around one carbon atom because the ort/ro-hydrogens clash 
with each other (they want to occupy the same space) and the compound actually adopts a propeller 
shape where each phenyl ring is slightly twisted relative to the next. 




the hydrogens 
in the ortho 
positions try 
to occupy the 
same space 




each phenyl ring is staggered relative to the next 



Even though complete derealization is not possible, each phenyl ring does lower the pK a because 
the sp carbon on the ring is electron-withdrawing. If we force the system to be planar, as in the com- 
pounds below, the p_K" a is lowered considerably. 



H H 





fluorene, pK a 22.8 
in the anion, the whole 
system is planar 



9-phenylfluorene, pK a 18.5 
in the anion, only the two 
fused fings can be planar 




fluoradene pK a 11 

in the anion, the whole 

system is planar 



The 'Fmoc' protecting group 

Sometimes in organic chemistry, when we are trying to do a 
reaction on one particular functional group, anothergroup in 
the molecule may also react with the reagents, often in a 
waythatwedonotwant. If a compound contains such a 
vulnerable group, we can 'protect' it by first converting it into 
a different less reactive group that can easily be converted 
back to the group that we want later. An example of such a 
'protecting group' is the Fmoc group used (for example, as 
the chloride, X = CI) to protect amines or alcohols. 




Fmoc-X 
Fmoc = 9-Fluorenylmethyloxycarbonyl 



The protecting group is removed using a base. This works 
because of the acidity of the proton in position 9 on the 
fluorene ring. Removal of that proton causes a breakup of 
the molecule with the release of the amine at the end. 




protected amine 



anion is stabilized by conjugation 
but can undergo elimination 



_y h 

NR, 



© 



HCT^NRo 






NHR, 



Fmoc-CI + NHR 2 



Fmoc NR 2 + HCI 



9-methylenefluorene 



C0 2 + NHR 2 

de-protected 



We saw in Chapter 7 how some compounds can become aromatic by gaining or losing electrons. 
Cyclopentadiene is one such compound, which becomes aromatic on deprotonation. The stability 
gained in becoming aromatic is reflected in the compound's p_K" a . 



196 



8 ■ Acidity, basicity, and pK a 




H H 



cyclopentadiene 
pK a 15.5 




anion is planar 

6ji electrons 

make it aromatic 




Me~ H 

cycloheptatriene 
pK a ca. 36 




anion not planar 
neither aromatic or anti-aromatic 



Me. 



,Me 



M£ H 

trimethylcyclopropene 
pK a ca. 62 



Me. 



,Me 



Me 

anion is planar 

4ji electrons 

make it anti-aromatic 



This is by no means as far as we can go. The five cyanide 
groups stabilize this anion so much that the pK a for this 
compound is about -11 and it is considerably more acidic 
than cyclopentadiene (pK a 15.5). 




The anions are also stabilized by 
solvation. Solvation is reduced by 
increasing the steric hindrance 
around the alkoxide. 



u 

X 



Compare the p_K" a of cyclopentadiene with that of cycloheptatriene. Whilst the anion 
of the former has 6 % electrons (which makes it isoelectronic with benzene), the anion 
of the latter has 8 7t electrons. Remember that on p. 176 we saw how An % electrons 
made a compound anti-aromatic? The cycloheptatrienyl anion does have 
An % electrons but it is not anti-aromatic because it isn't planar. However, it certainly 
isn't aromatic either and its p-ST a of around 36 is about the same as that of propene. This 
contrasts with the cyclopropenyl anion, which must be planar since any three points 
define a plane. Now the compound is anti-aromatic and this is reflected in the very high 
piC a (about 62). Other compounds may become aromatic on losing a proton. We 
looked at fluorene a few pages back: now you will see that fluorene is acidic because its 
anion is aromatic (14 % electrons). 



# The more stabilized the conjugate base, A , the stronger is the acid, HA. Ways to 
stabilize A~ include: 

• Having the charge on an electronegative element 

• Delocalizing the negative charge over other carbon atoms, or even better, over 
more electronegative atoms 

• Spreading out the charge over electron-withdrawing groups by the polariza- 
tion of a bonds (inductive) 

• Having the negative charge in an orbital with more s character 

• Becoming aromatic 

Electron-donating groups decrease acidity 

All of the substituents in the examples above have been electron-withdrawing and have helped to sta- 
bilize the negative charge of the conjugate base, thereby making the acid stronger. What effect would 
electron-donating groups have? As you would expect, these destabilize the conjugate base because, 
instead of helping to spread out the negative charge, they actually put more in. The most common 
electron-donating groups encountered in organic chemistry are the alkyl groups. These are weakly 
electron-releasing (p. 416). 






CH 3 



H 3 C OH 



H 3 C 



OH 



H ^OH 


H 3 C^ ^OH 




rmic (methanoic) acid 


acetic (ethanoic) acid 


methanol 


pK a 3.7 


pK a 4.8 


pK a 15.5 



H,C 



OH 



ethanol 
pK a 16.0 



isopropyl alcohol 
PKa 17.1 



CH 3 

H 3 C>'^^ 
H 3 (T OH 

tert-butyl alcohol 
pK a 19.2 



Basicity 



197 



Although to a lesser extent than amides (p. 165), the ester group is also stabilized by conjugation. In 
this case, the 'ethoxide part' of the ester is electron-releasing. This explains the p-JC a s shown below. 




H H 



H 3 C 




EtO 




H H 3 C 



H H 




EtO 




H H 



EtO 




OEt 



acetaldehyde 


acetone 


ethyl acetate 


propandial 


acetylacetone 


ethyl acetoacetate 


diethyl malonate 


(ethanal) 


(propanone) 


(ethyl ethanoate) 


pK a ca. 5 


(2,4-pentanedione) 


(ethyl 3-oxobutanoate) 


(diethyl propanedioate) 


pK a 13.5 


pK a 20 


pK a 25 




pK a 8.9 


pK a 10.6 


pK a 12.9 



Nitrogen acids 

Oxygen acids and carbon acids are by far the most important examples you will encounter and 
by now you should have a good understanding of why their piC a values are what they are. Before 
we move on to bases, it would be worthwhile to remind you how different nitrogen acids are from 
oxygen acids, since the conjugate bases of amines are so important. The p_K" a of ammonia is much 
greater than the p_K" a of water (about 33 compared with 15.74). This is because oxygen is more 
electronegative than nitrogen and so can stabilize the negative charge better. A similar trend 
is reflected in the pK a s of other nitrogen compounds, for example, in the amide group. Whilst the 
oxygen equivalent of an amide (a carboxylic acid) has a low piC a , a strong base is needed to deproto- 
nate an amide. Nevertheless, the carbonyl group of an amide does lower the p-K" a from that of 
an amine (about 30) to around 17. It's not surprising, therefore, that the two carbonyl groups in 
an imide lower the pK a still further, as in the case of phthalimide. Amines are not acidic, amides 
are weakly acidic (about the same as alcohols), and imides are definitely acidic (about the same as 
phenols). 



N — H 











3 

1 


f 


V\ 


H^H 





H V H 


j 




k 


H 


water 


carboxylic acid 


ammonia 


amide 




phthalimide 


pK a 15.74 


pK a ca. 5 


pK a ca. 33 


pK 


a ca. 17 




pK a 8.3 



Basicity 

A base is a substance that can accept a proton by donating a pair of electrons. We have 
already encountered some — for example, ammonia, water, the acetate anion, and the methyl 
anion. The question we must now ask is: how can we measure a base's strength? To what extent 
does a base attract a proton? We hope you will realize that we have already addressed this prob- 
lem by asking the same question from a different viewpoint: to what extent does a protonated 
base want to keep its proton? For example if we want to know which is the stronger base — formate 
anion or acetylide anion — we look up the pK a s for their conjugate acids. We find that the pJC a 
for formic acid (HCO2H) is 3.7, whilst the pK a for ethyne (acetylene) is around 25. This means 
that ethyne is much more reluctant to part with its proton, that is, acetylide is much more basic 
than formate. This is all very well for anions — we simply look up the piC a value for the neutral conju- 
gate acid, but what if we want to know the basicity of ammonia? If we look up the pK a for ammonia 
we find a value around 33 but this is the value for deprotonating neutral ammonia to give the amide 
ion, NH 2- 



The potassium salt of 6-methyl-l,2,3- 
oxathiazin-4-one 2,2-dioxide known as 
acesulfame-K is used as an artificial 
sweetener (trade name Sunett). Here 
the negative charge is delocalized over 
both the carbonyl and the sulfone 
groups. 



Me 




acesulfame-K 



► Amides 

Do not get confused with the two 
uses of the word 'amide' in 
chemistry. Both the carbonyl 
compound and the 'ionic' base 
formed by deprotonating an amine 
are known as amides. From the 
context it should be clear which is 
meant — most of the time chemists 
(at least organic chemists!) mean 
the carbonyl compound. 

u 

the amide group 

u 

I base 

R 1 ^ ^R 2 R 1 ^ N* 2 



an amine 



an amide base 

e.g. sodium amide 

NaNH ? 



198 



8 ■ Acidity, basicity, and pK a 



Get a feel for pK a s! 

Remember that the pK a also 
represents the pH when we have 
equal concentrations of acid and 
conjugate base, that is, NH3 and 
NHj in this case. You know that 
ammonia is a weak base and that 
an aqueous solution is alkaline so 
it should come as no surprise that 
its pK a is on the basic side of 7. 
To be exact, at pH 9.24 an 
aqueous solution of ammonia 
contains equal concentrations of 
ammonium ions and ammonia. 



If we want to know the basicity of ammonia, we must look up the p_fC a of its conjugate acid, the 
ammonium cation, NH4, protonated ammonia. Its piC a is 9.24 which means that ammonia is a 
weaker base than hydroxide — the p_K" a for water (the conjugate acid of hydroxide) is 15.74 (p. 190). 
Now we can summarize the states of ammonia at different pH values. 



NH® ^r NH 3 NH 3Z= ^NH® 




pH <9.24 pH 7 pH 9.24 


pH 10-33 pH -33 pH >33 




neutral to acid 


strongly basic very strongly basic 


nde 


pH 16 
limit of measurements in water 





Scales for basicity — pK B and pK aH 

The material in this box is quite mathematical and may be 
skipped if you find it too alien. 

It is often convenient to be able to refer to the basicity of a 
substance directly. In some texts a different scale is 
used, pKb. This is derived from considering how much 
hydroxide ion a base forms in water ratherthan how much 
hydronium ion the conjugate acid forms. 

For the pK B scale: 

B(aq) + H 2 ^=^ OH(aq) + BH + (aq) 



Kr 



[OhT][BH + ] 
El 



Hence 

pK B = -log(K B ) 
For the pK a scale: 

BH + (aq) + H 2 ^= H 3 + (aq) + B(aq) 



Ka 



[H 3 + ][B] 
[BH + ] 
Hence 

pK A = -log(K A ) 



Just as in the acid pK a scale, the lower the pK a the 
stronger the acid, in the basic pK B scale, the lower the 
pK B , the strongerthe base. The two scales are related: 
the product of the equilibrium constants simply equals the 
ionic product of water. 



K B xK* 



[0H-][BH + ] ;; [H 3 + ][B] 
[B] X [BH + ] 

= [0H-][H 3 + ] = K w =l(r 



that is, 



pK A + pK B = pK w = 14 

There is a separate scale for bases, but it seems silly to 
have two different scales, the basic pK B and the familiar 
pK a , when one will do and so we will stick to pK a . 
However, to avoid any misunderstandings that can arise 
from amphoteric compounds like ammonia, whose pK a is 
around 33, we will either say: 

• The pK a of ammonia's conjugate acid is 9.24 
or, more concisely, 

• The pK a H of ammonia is 9.24 (where pK a H simply 
means the pK a of the conjugate acid) 



The most important factor in the 
strength of a base is which 
element the lone pair (or negative 
charge) is on. The more 
electronegative the element, the 
tighter it keeps hold of its 
electrons, and so the less 
available they are to accept a 
proton, and the weaker is the 
base. 



What factors affect how basic a compound is? 

This really is the same as the question we were asking about the strength of an acid — the more 'sta- 
ble' the base, the weaker it is. The more accessible the electrons are, the stronger the base is. 
Therefore a negatively charged base is more likely to pick up a proton than a neutral one; a com- 
pound in which the negative charge is delocalized is going to be less basic than one with a more con- 
centrated, localized charge, and so on. We have seen that carboxylic acids are stronger acids than 
simple alcohols because the negative charge formed once we have lost a proton is delocalized over 
two oxygens in the carboxylate but localized on just one oxygen for the alkoxide. In other words, the 
alkoxide is a stronger base because its electrons are more available to be protonated. Since we have 
already considered anionic bases, we will now look in more detail at neutral bases. 

A 



B: 



.H 



H 



© 
B — H 



© 
— H 



There are two main factors that determine the strength of a neutral base: how accessible is the 
lone pair and to what extent can the resultant positive charge formed be stabilized either by dereal- 
ization or by the solvent. The accessibility of the lone pair depends on its energy — it is usually the 
HOMO of the molecule and so, the higher its energy, the more reactive it is and hence the stronger 
the base. The lone pair is lowered in energy if it is on a very electronegative element or if it can be 
delocalized in some manner. 



Neutral nitrogen bases 



199 



This explains why ammonia is 10 times more basic than water: since oxygen is more electroneg- 
ative than nitrogen, its lone pair is lower in energy. In other words, the oxygen atom in water wants 
to keep hold of its electrons more than the nitrogen in ammonia does and is therefore less likely to 
donate them to a proton. The P-K" a H for ammonia (that is, the pK a for ammonium ion) is 9.24 whilst 
the pJC[H for water (the piC a for hydronium ion) is -1.74. Nitrogen bases are the strongest neutral 
bases commonly encountered by the organic chemist and so we will pay most attention to these in 
the discussion that follows. 



► P«a H ! 






We use pK aH to mean 


the pK a 


of 


the conjugate acid. 







Neutral nitrogen bases 



R 

Me 

Et 
n-Pr 



n-Bu 10.7 



Ammonia is the simplest nitrogen base and has a piC a H of 9.24. Any substituent that increases the 
electron density on the nitrogen therefore raises the energy of the lone pair thus making it more 
available for protonation and increasing the basicity of the amine (larger p-K" a H)- Conversely, any 
substituent that withdraws electron density from the nitrogen makes it less basic (smaller pK a n). 

Effects that increase the electron density on nitrogen 

We can increase the electron density on nitrogen either by attaching an electron-releas- 

ing group or by conjugating the nitrogen with an electron-donating group. The simplest 
example of an electron-releasing group is an alkyl group (p. 416). If we successively sub- 
stitute each hydrogen in ammonia by an electron-releasing alkyl group, we should 
increase the amine's basicity. The piC a H values for various mono-, di-, and trisubstituted 
amines are shown in Table 8.4. 
Points to notice in Table 8.4: 

• All the amines have P-KaHS greater than that of ammonia (9.24) 

• All the primary amines have approximately the same P-^h (about 10.7) 

• All the secondary amines have p^ a H s that are slightly higher 

• Most of the tertiary amines have pK a ^s lower than those of the primary amines 

The first point indicates that our prediction that replacing the hydrogens by electron-releasing alkyl 
groups would increase basicity was correct. A strange feature though is that, whilst substituting one 
hydrogen of ammonia increases the basicity by more than a factor often (one p_K" a unit), substituting two 
has less effect and in the trisubstituted amine the p-K" a n is actually lower. So far we have only considered 
one cause of basicity, namely, the availability of the lone pair but the other factor, the stabilization of the 
resultant positive charge formed on protonation, is also important. Each successive alkyl group does 
help stabilize the positive charge because it is electron-releasing but there is another stabilizing effect — 
the solvent. Every hydrogen attached directly to nitrogen will be hydrogen bonded with solvent water 
and this also helps to stabilize the charge: the more hydrogen bonding, the more stabilization. The 
observed basicity therefore results from a combination of effects: ( 1 ) the increased availability of the lone 
pair and the stabilization of the resultant positive charge, which increases with successive replacement of 
hydrogen atoms by alkyl groups; and (2) the stabilization due to solvation, an important part of which is 
due to hydrogen bonding and this effect decreases with increasing numbers of alkyl groups, 
more stabilization of positive charge from alkyl groups _ .... 

^ Gas phase acidity 



Table 8.4 pK a n values for primary, secondary, 
and tertiary amines 



P«aH RNH 2 

10.6 

10.7 
10.7 














■V'H 



© 



T R 



© 



N£"R 

^R 



If we look at the pK aH values in the gas 
phase, we can eliminate the hydrogen 
bonding contribution and we find the 
basicity increases in the order we expect, 
that is, tertiary > secondary > primary. 



more stabilization of positive charge from hydrogen bonding with solvent 



pK aH R 2 NH 

10.8 

11.0 
11.0 
11.3 



P«aH R 3 N 

9.8 

10.8 
10.3 

9.9 



Introducing alkyl groups is the simplest way to increase the electron density on nitrogen but there 
are other ways. Conjugation with an electron- donating group produces even stronger bases (p. 202) 
but we could also increase the electron density by using elements such as silicon. Silicon is more 



200 



8 ■ Acidity, basicity, and pK a 



Remember that Me 4 Si, tetramethyl 
silane, in carbon NMR resonates at 0.0 
p. p.m.? This is another consequence of 
silicon being more electropositive than 
carbon — the methyl carbons of TMS are 
more shielded and so resonate at a 
smaller chemical shift than other 
saturated carbons. 



Compare this summary with the one for 
the stabilization of the conjugate base 
A~on p. 196. In both cases, we are 
considering the same factors. 



Me 




Me""i' 



Me>»i 




pK aH 11.0 



electropositive than carbon, that is, it pushes more electron 
density on to carbon. This extra donation of electrons also 
means that the silicon compound has a higher p_K" a H value 
than its carbon analogue since the nitrogen's lone pair is 
higher in energy. 



• Effects that decrease the electron density on nitrogen 

The lone pair on nitrogen will be less available for protonation, and the amine less 
basic, if: 

• The nitrogen atom is attached to an electron-withdrawing group 

• The lone pair is in an sp or sp hybridized orbital 

• The lone pair is conjugated with an electron-withdrawing group 

• The lone pair is involved in maintaining the aromaticity of the molecule 

The p^aH s °f some amines in which the nitrogen is attached either directly or indirectly to an 
electron- withdrawing group are shown below. We should compare these values with typical values 
of about 1 1 for simple primary and secondary amines. 



Of course, electron-withdrawing groups 
on the benzene ring will affect the 
availability of the lone pair. For 
example, the pK aH of p^nitro aniline is 
only 1.11. This explains why certain 
aromatic amines (for example, 
nitroanilines and dibromoanilines) 
can't be acetylated using the 
Lumiere-Barbier method (p. 188). 



CI 3 C- -NH 2 C I 3 C-^^ 2 F 3° NH 2 F 3 C 

pK aH 5.5 pK aH 9.65 pK aH 5.7 pK aH 8.7 

The strongly electron-withdrawing CF 3 and CCI3 groups have a large effect when they are on the 
same carbon atom as the NH 2 group but the effect gets much smaller when they are even one atom 
further away. Inductive effects fall off rapidly with distance. 



Hybridization is important 

As explained on p. 194, the more s character an orbital 
has, the more tightly it holds on to its electrons and so the 
more electron-withdrawing it is. This is nicely illustrated by 
the series in Table 8.5. 

• These effects are purely inductive electron withdrawal. 
Satisfy yourself there is no conjugation possible 

• The last compound's pK a H is very low. This is even less 
basic than a carboxylate ion. 



Table 8.5 pK a ns of unsaturated 
primary, secondary, and tertiary amines 

R RNH 2 R 2 NH R 3 N 

H 3 C— CH 2 — CH 2 — 10.7 11.0 10.3 



H 2 C=CH— CH 2 



9.5 



9.3 



8.3 



HC=C — CH 2 — 



8.2 



6.1 



3.1 



If the lone pair itself is in an sp or an sp orbital, it is more tightly held (the orbital is lower in ener- 
gy) and therefore much harder to protonate. This explains why the lone pair of the nitrile group is 
not at all basic and needs a strong acid to protonate it. 





Me- 



ENO 



lone pair in sp 3 orbital 
pK aH 10.7 



lone pair in sp 2 orbital 
pK aH 9.2 



H H 

lone pair in sp 3 orbital 
pK aH 10.8 



lone pair in sp orbital 
pK aH ca. -10 



The low p-K" a H of aniline (PI1NH2), 4.6, is partly due to the nitrogen being attached to an sp car- 
bon but also because the lone pair can be delocalized into the benzene ring. In order for the lone pair 
to be fully conjugated with the benzene ring, the nitrogen would have to be sp hybridized with the 
lone pair in the p orbital. This would mean that both hydrogens of the NH2 group would be in the 
same plane as the benzene ring but this is not found to be the case. Instead, the plane of the NH 2 
group is about 40° away from the plane of the ring. That the lone pair is partially conjugated into the 
ring is shown indirectly by NMR shifts and by the chemical reactions that aniline undergoes. Notice 



Neutral nitrogen bases 



201 



that, when protonated, the positive charge cannot be delocalized over the benzene ring and any sta- 
bilization derived from the lone pair in unprotonated aniline being delocalized into the ring is lost. 




cyclohexylamine 
pK aH 10.7 




aniline 
pK aH 4.6 




o\»"J 



ca. 40° 



the NH 2 group is about 40° away 
from being in the plane of the ring 



Amides are weak bases protonated on oxygen 

In contrast to aromatic amines, the amide group is completely planar (p. 165) with the nitrogen sp 
hybridized and its lone pair in the p orbital, thereby enabling it to overlap effectively with the car- 
bonyl group. 



1 





nitrogen is sp 2 hybridized with 
its lone pair in a p orbital 



good overlap 
with the carbonyl group 



derealization of nitrogen's 
lone pair into % system 



This delocalization 'ties up' the lone pair and makes it much less basic: the p-K" a H for an amide 
is typically between and — 1. Because of the delocalization amides are not protonated on 
nitrogen. 



H-Qv 



.© 



/' 






H H 



no protonation occurs on the nitrogen atom 

Protonation at nitrogen would result in a positive charge on the nitrogen atom. Since this is adja- 
cent to the carbonyl, whose carbon is also electron-deficient, this is energetically unfavourable. 
Protonation occurs instead on the carbonyl oxygen atom. We can draw the mechanism for this using 
either a lone pair on oxygen or on nitrogen. 



o- 



A„A 



A,.- 



protonation occurs 
on the oxygen atom 




these structures are just two different 
ways to draw the same delocalized cation 



these arrows emphasise 

the contribution of the 

nitrogen's lone pair. 



Furthermore, if the amide were protonated at 
nitrogen, the positive charge could not be delocalized 
on to the oxygen but would have to stay localized on 
the nitrogen. In contrast, when the amide is protonat- 
ed on the oxygen atom, the charge can be delocalized 
on to the nitrogen atom making the cation much 
more stable. We can see this if we draw delocalization 
arrows on the structures in the green box. 




202 



8 ■ Acidity, basicity, and pK a 



<&. 



X X 



an amidine 



pK aH 12.4 



(+) 




(-) 


NH 2 






ii 


■ 


R 


■ 


(+) 




(-) 


amidinium 


ca 


rboxylate 


cation 




anion 



OS /H 

N 

X 



guanidine: pK a 13.6 




(+)H 2 N<^' ^NH 2 (+) 



very stable guanidinium cation 
each (+) is a third 
of a positive charge 



(-)O^ '>0 (-) 

very stable carbonate dianion 

each (-) is two-thirds 

of a negative charge 



Amidines are stronger bases than amides or amines 

An amidine is the nitrogen equivalent of an amide — a C=NH group replaces the carbonyl. Amidines 
are much more basic than amides, the p_K" a HS of amidines are larger than those of amides by about 13 
so there is an enormous factor of 10 in favour of amidines. In fact, they are among the strongest 
neutral bases. 

An amidine has two nitrogen atoms that could be protonated — one is sp hybridized, the other 
sp hybridized. We might expect the sp nitrogen to be more basic but protonation occurs at the sp 
nitrogen atom. This happens because we have the same situation as with an amide: only if we proto- 
nate on the sp nitrogen can the positive charge be delocalized over both nitrogens. We are using 
both lone pairs when we protonate on the sp nitrogen. 



A^-H O./H 







iP-H 



s" 




NH 3 



The electron density on the sp nitrogen in an amidine is increased through conjugation with the 
sp nitrogen. The delocalized amidinium cation has identical C-N bond lengths and a positive 
charge shared equally between the two nitrogen atoms. It is like a positively charged analogue of the 
carboxylate ion. 



Amidine bases 


<^\ 


n 


Two frequently used amidine bases are 


X J 


rr 


DBN (l,5-diazatacyclo[3.4.0]nonene-5) and 


<T 


DBU(l,8-diaza£>icyclo[5.4.0]undecene-7). 


\J 


\ ) 


They are easier to make, more stable, and less 




\ — / 


volatile than simpler amidines. 


DBN 


DBU 



Guanidines are very strong bases 

Even more basic is guanidine, p-fC a H 13.6, nearly as strong a base as NaOH! On protonation, the pos- 
itive charge can be delocalized over three nitrogen atoms to give a very stable cation. All three nitro- 
gen lone pairs cooperate to donate electrons but protonation occurs, as before, on the sp nitrogen 
atom. 



jPlhO^h 



A. 



H,I\T ^NH 




/P-H 



H 2 N 




NH 2 



This time the resulting guanidinium ion can be compared to the very stable carbonate dianion. 
All three C-N bonds are the same length in the guanidinium ion and each nitrogen atom has the 
same charge (about one-third positive). In the carbonate dianion, all three C-O bonds are the same 
length and each oxygen atom has the same charge (about two-thirds negative as it is a dianion). 

Imidazoline is a simple cyclic amidine and its pi^H value is just what we expect, around 11. 
Imidazole, on the other hand, is less basic (pi^H 7.1) because both nitrogens are attached to an 
electron-withdrawing sp carbon. However, imidazole, with its two nitrogen atoms, is more basic 
than pyridine (p-K" a H 5-2) because pyridine only has one nitrogen on which to stabilize the positive 
charge. 

r\ i=\ 

V/ NH 



imidazoline 



V H ^ 


HN^SyNH 


imidazole 


imidazolium 


pK aH 7.1 


cation 




pyridine 
pK aH 5.2 




Neutral oxygen bases 



203 



Both imidazole and pyridine are aromatic — they are flat, cyclic molecules with 6 71 electrons in the 
conjugated system (p. 177). Imidazole has one lone pair that is and one that is not involved in the 
aromaticity (Chapter 43). 



this lone pair is in an sp 2 orbital 

and is not involved with the aromaticity 

of the ring. Protonation occurs here 



the aromaticity of imidazole 



this lone pair is in a 
p orbital contributing 
to the 6ti electrons 
in the aromatic ring 



Protonation occurs on the nitrogen atom having the sp lone pair because both lone pairs con- 
tribute and the resulting delocalized cation is still aromatic. Pyridine is also protonated on its sp 
lone pair (it is the only one it has!) and the pyridinium ion is also obviously aromatic — it still has 
three conjugated 7t bonds in the ring. 

aromatic imidazole 






HI\K .NH 



aromatic imidazoium ion 



This contrasts to pyrrole in which the lone pair on the only nitrogen atom is needed to complete 
the six aromatic 7t electrons and is therefore delocalized around the ring. Protonation, if it occurs at 
all, occurs on carbon rather than on nitrogen since the cation is then delocalized. But the cation is no 
longer aromatic (there is a saturated CH2 group interrupting the conjugation) and so pyrrole is not 
at all basic (p-K" a H about -4). 



o 

H 

pyrrole 
pK aH ca. -4 




this lone pair is in a 
p orbital contributing 
to the 6ji electrons 
in the aromatic ring 



^-«^_ 



the aromaticity of pyrrole 



N 

I 
H 

aromatic pyrrole 




nonaromatic cation 



Neutral oxygen bases 



We have already seen that water is a much weaker base than ammonia because oxygen is more elec- 
tronegative and wants to keep hold of its electrons (p. 199). Oxygen bases in general are so much 
weaker than their nitrogen analogues that we don't regard them as bases at all. It is still important to 
know the p-ST a HS of oxygen compounds because the first step in many acid-catalysed reactions is pro- 
tonation at an oxygen atom. Table 8.6 gives a selection of piC^s of oxygen compounds. 



Table 8.6 pK aH s 
Oxygen compound 


of oxygen compounds 

Oxygen compound 
(conjugate base A) 


Approximate pK aH 
of oxygen compound 
(pK a of acid HA) 


Conjugate acid HA of oxygen 
compound 


ketone 




R^R 


-7 




© OH 

X 


carboxylic acid 




x 


-7 




R^^OH 


phenol 


OH 

6 


-7 




© 0H 2 

6 


carboxylic ester 




x 

R^^OR 


-5 




© OH 

x 

R'^^OR 



204 



8 ■ Acidity, basicity, and pK a 



Table 8.6 (continued) 

Oxygen compound Oxygen compound 

(conjugate base A) 



Approximate pK aH 
of oxygen compound 
(pK a of acid HA) 



Conjugate acid HA of oxygen 
compound 



alcohol 



ether 



R- -H 



R-% 



R © H 



r^©S* 



water 



amide 



H^H 





x 



u 

X 



-1.74 



-0.5 



H ©^H 



un 



© 
NR 2 



All the same factors of electron donation and withdrawal apply to oxygen compounds as well as to 
nitrogen compounds, but the effects are generally much less pronounced because oxygen is so elec- 
tronegative. In fact, most oxygen compounds have piC^s around -7, the notable exception being 
the amide, which, because of the electron donation from the nitrogen atom, has a p-K" a H around -0.5 
(p. 201). They are all effectively nonbasic and strong acids are needed to protonate them. 



Histamine in this example is an 
agonist in the production of 
gastric acid. It binds to specific 
sites in the stomach cells 
(receptor sites) and triggers the 
production of gastric acid (mainly 
HCI). 

An antagonist works by binding to 
the same receptors but not 
stimulating acid secretion itself. 
This prevents the agonist from 
binding and stimulating acid 
production. 



pK a in action — the development of the drug cimetidine 



< 



Me 




,/ 



CN 



JL .Me \ 1 



cimetidine 



histamine 



The development of the anti-peptic ulcer drug cimetidine gives a fascinating insight into the impor- 
tant role of p_K" a in chemistry. Peptic ulcers are a localized erosion of the mucous membrane, resulting 
from overproduction of gastric acid in the stomach. One of the compounds that controls the produc- 
tion of the acid is histamine. 
(Histamine is also responsible for 
the symptoms of hay fever and 
allergies.) 

Histamine works by binding 
into a receptor in the stomach lin- 
ing and stimulating the produc- 
tion of acid. What the developers of cimetidine at SmithKline Beecham wanted was a drug that 
would bind to these receptors without activating them and thereby prevent histamine from binding 
but not stimulate acid secretion itself. Unfortunately, the antihistamine drugs successfully used in 
the treatment of hay fever did not work — a different histamine receptor was involved. Notice that 
cimetidine and histamine both have an imidazole ring in their structure. This is not coincidence — 
cimetidine's design was centred around the structure of histamine. 

In the body, most histamine exists as a salt, being protonated on the primary amine and the early 
compounds modelled this. The guanidine analogue was synthesized and tested to see if it had any 
antagonistic effect (that is, if it could bind in the histamine receptors and prevent histamine bind- 
ing). It did bind but unfortunately 
it acted as an agonist rather than 
an antagonist and stimulated acid 
secretion rather than blocking it. 
Since the guanidine analogue has 
a p-JC a H even greater than hista- 
mine (about 14.5 compared to 
about 10), it is effectively all pro- 
tonated at physiological pH. 



pK a 10 



14.5 







© 

NH 3 



the major form of histamine 
at physiological pH (7.4) 




the guanidine analogue 

the extra carbon in the chain was found 

to increase the efficacy of the drug 



pK a in action — the development of the drug cimetidine 



205 



The agonistic behaviour of the drug clearly had to be suppressed. The thought occurred to the 
SmithKline Beecham chemists that perhaps the positive charge made the compound agonistic, and so 
a polar but much less basic compound was sought. Eventually, they came up with burimamide. The 
most important change is the replacement of the C=NH in the guanidine compound by C=S. Now 
instead of a guanidine we have a thiourea which is much less basic. (Remember that amidines, p. 202, 




introduction of sulfur decreased the 
pK a to ca. -1 so now this group is 
no longer protonated 



are very basic but that amides burimamide 
aren't? The thiourea is like the 
amide in that the sulfur withdraws 
electrons from the nitrogens.) The 
other minor adjustments, increas- 
ing the chain length and adding the 
methyl group on the thiourea, fur- 
ther increased the efficacy. 

The new compound was a fairly good antagonist (that is, bound in the receptors and blocked his- 
tamine) but more importantly shown no agonistic behaviour at all. The compound was such a 
breakthrough that it was given a name, 'burimamide', and even tested in man. Burimamide was 
good, but unfortunately not good enough — it couldn't be given orally. A rethink was needed and 
this time attention was focused on the imidazole ring. 



extra chain length and the methyl group increased the activity still further 



positive charge here withdraws electrons 
and decreases pK aH of ring 



9 iX^ 



© 

NH 3 



imidazole histamine 

pK aH 6.8 pK aH of imidazole ring 5.9 



thiourea too far away from ring to influence pK aH 
alkyl chain is electron-donating and raises pK aH of ring 



Me 




H H 

burimamide 
pK aH of imidazole ring 7.25 



When the drug was invented, the 
company was called Smith, Kline, 
and French (SKF) but after a 
merger with Beechamsthe 
company is now called SmithKline 
Beecham or SB. Things may have 
changed further by the time you 
read this book. 



The pXaH of the imidazole ring in burimamide is significantly greater than that in histamine: the 
longer alkyl group in burimamide is electron-donating and raises the P-JQh of the ring. In histamine, 
on the other hand, the positive charge of the protonated amine withdraws electrons and decreases 
the P-K" a H- This means, of course, that there will be a greater proportion of protonated imidazole 
(imidazolium cation) in burimamide and this might hinder effective binding in the histamine recep- 
tor site. So the team set out to lower the p-K" a H of the imidazole ring. It was known that a sulfur occu- 
pies just about the same space as a methylene group, -CH2-, but is more electron-withdrawing. 
Hence 'thiaburimamide' was synthesized. 





tautomers of thiaburimamide: pK aH of imidazole ring 6.25 

In turns out that one tautomer of the imidazole ring binds better than the other (and much better 
than the protonated form). The introduction of a methyl group on the ring was found to increase the 
proportion of this tautomer and did indeed improve binding to the histamine receptor, even though 
the p-K" a H of the ring was raised because of the electron- donating character of the methyl group. 

H 



Tautomers are isomers differing only in 
the positions of hydrogen atoms and 
electrons. Otherwise the carbon 
skeleton is the same. They will be 
explained in Chapter 21. 






^/ 






_< 



Me 




Me 



A B 

these two tautomers are in rapid equilibrium 
we want tautomer 'A' 



introduction of an electron-releasing group 
favoured tautomer 'A' 



metiamide: pK aH of imidazole ring 6.8 



206 8 ■ Acidity, basicity, and pK a 



The new drug, metiamide, was ten times more effective than burimamide when tested in man. 
However, there was an unfortunate side-effect: in some patients, the drug caused a decrease in the 
number of white blood cells, leaving the patient open to infection. This was eventually traced back to 
the thiourea group. The sulfur had again to be replaced by oxygen, to give a normal urea and, just to 
see what would happen, by nitrogen to give another guanidine. 

H H 

N ^ Me 

o / J^ NH 



,Me 




urea analogue of metiamide guanidine analogue of metiamide 

Neither was as effective as metiamide but the important discovery was that the new guanidine no 
longer showed the agonistic effects of the earlier guanidine. Of course, the guanidine would also be 
protonated so we had the same problem we had earlier — how to decrease the p-KaH of the guanidine. 

A section of this chapter considered the effect of electron-withdrawing groups on piC a H and 
showed that they reduce the piC a H and make a base less basic. This was the approach now adopted — 
the introduction of electron-withdrawing groups on to the guanidine to lower its p^ a H- Table 8.7 
shows the p-K" a H s of various substituted guanidines. 

HolNr^NH, 



Clearly, the cyano- and nitro-substituted guanidines would not be protonated at all. These were 
synthesized and found to be just as effective as metiamide but without the nasty side-effects. Of the 
two, the cyanoguanidine compound was slightly more effective and this was developed and named 
'cimetidine'. 

H 

N 



Table 8.7 pK a ns of substituted guanidines 








/ R 

N 


R H Ph CH3CO NH 2 C0 MeO 


CN 


N0 2 




1© ~ 


pK aH 14.5 10.8 8.33 7.9 7.5 


-0.4 


-0.9 


H 2 N^ 


^-T\IH 2 




A. 



.Me 
N 
H H 

the end result, cimetidine 

The development of cimetidine by Smith, Kline, and French from the very start of the project up 
to its launch on the market took thirteen years. This enormous effort was well rewarded — Tagamet 
(the trade name of the drug cimetidine) became the best-selling drug in the world and the first to 
gross more than one billion dollars per annum. Thousands of ulcer patients worldwide no longer 
had to suffer pain, surgery, or even death. The development of cimetidine followed a rational 
approach based on physiological and chemical principles and it was for this that one of the scientists 
involved, Sir James Black, received a share of the 1988 Nobel Prize for Physiology or Medicine. None 
of this would have been possible without an understanding of p_fC a s. 



Problems 



207 



Problems 

1. If you wanted to separate a mixture of naphthalene, pyridine, 
and p-toluic acid, how would you go about it? All three 
compounds are insoluble in water. 

^C0 2 H 






naphthalene 



pyridine 



para-toluic acid 



2. In the separation of benzoic acid from toluene we suggested 
using NaOH solution. How concentrated a solution would be 
necessary to ensure that the pH was above the p_K" a of benzoic acid 
(p.K" a 4.2)? How would you estimate how much solution to use? 

3. What species would be present if you were to dissolve this 
hydroxy-acid in: (a) water at pH 7; (b) aqueous alkali at pH 12; or 
(c) a concentrated solution of a mineral acid? 




4. What would you expect to be the site of (a) protonation and (b) 
deprotonation if the compounds below were treated with an 
appropriate acid or base. In each case suggest a suitable acid or 
base for both purposes. 





.0© 

+ 



(JJ + A 0H 



Art /= \ 
(b) HN^NH 



(b) 










,0 





6. What is the relationship between these two molecules? Discuss 
the structure of the anion that would be formed by the 
deprotonation of each compound. 



OH 



7. What species would be formed by treating this compound 
with: (a) one equivalent; (b) two equivalents of NaNH 2 in liquid 
ammonia? 



8. The carbon NMR spectra of these compounds could be run in 
D 2 under the conditions shown. Why were these conditions 
necessary and what spectrum would you expect to observe? 



H 2 N / 




C spectrum run in DC1/D 2 




0^ ^ ^OH 

13 C spectrum run in NaOD/D 2 



9. The phenols shown here have approximate piC a values of 4, 7, 9, 

10, and 11. Suggest with explanations which p_K" a value belongs to 
which phenol. 

,OH ^>^ .OH 



0,N 




5. Suggest what species would be formed by each of these |y| e 
combinations of reagents. You are advised to use p_K" a values to 
help you and to beware of some cases where 'no change' might be 
the answer. "■ 




N0 2 2 N 

OH ^^ ^OH 

Me 






10. Discuss the stabilization of the anions formed by the 
deprotonation of (a) and (b) and the cation formed by the 
protonation of (c). Consider derealization in general and the 
possibility of aromaticity in particular. 
(a) (b) o 




\ / 



// 



(c) 



I 





%. y^ 



rv 



208 



8 ■ Acidity, basicity, and pK a 



11. The pK a values for the amino acid cysteine are 1.8, 8.3, and 14. Neither of these methods of making pentan-l,4-diol will 
10.8. Assign these pK a values to the functional groups in cysteine work. Explain why not — what will happen instead? 



and draw the structure of the molecule in aqueous solution at the 
following pHs: 1, 5, 9, and 12. 



HS 



C0 2 H 



NH 2 

12. Explain the variations in the p_K" a values for these carbon acids. 


XX ^„ ^ 0E , 



pK a 9 



pK a 5.9 



pK a 10.7 




A^» A^ AA, 



pK a 16.5 



pK a 5.1 



pK a 4.7 



Me — MgBr 



OH 




Mg 



Et 2 



BrM£ 




13. Explain the various pK a values for these derivatives of the 
naturally occurring amino acid glutamic acid. Say which pJC a 
belongs to which functional group and explain why they vary in 
the different derivatives. 



C0 2 H 



HO, 

NH 2 

glutamic acid; pK a s 2.19, 4.25, and 9.67 

C0 2 H Et0 2 C 



H 2 N0 

NH 2 

glutamine; pK a s 2.17 and 9.13 



C0 2 Et 



NH 2 

diethyl ester; pK a 7.04 



EtO; 



C0 2 H 



HO- 



NH 2 

monoethyl ester; 
pK a s 2.15 and 9.19 



C0 2 Et 



NH 2 

monoethyl ester; 
pK a s 3.85 and 7.84 



Using organometallic reagents 
to make C-C bonds 



9 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• Electronegativity and the polarization 
of bonds ch4 


• Organometallics: nucleophilic and 
often strongly basic 


• More about organometallics chlO & 
ch48 


• Grignard reagents and organolithiums 
attack carbonyl groups ch6 


• Making organometallics from halo- 
compounds 


• More ways to make C-C bonds from 
C=0 groups ch26-ch29 


• C-H deprotonated by very strong 
bases ch8 


• Making organometallics by 
deprotonating carbon atoms 

• Using organometallics to make new 
C-C bonds from C=0 groups 


• Synthesis of molecules ch 25 &ch30 



Introduction 

In Chapters 2-8 we covered basic chemical concepts, which mostly fall under the headings 'struc- 
ture' (Chapters 2-4 and 7) and 'reactivity' (Chapters 5, 6, and 8). These concepts are the bare bones 
supporting all of organic chemistry, and now we shall start to put flesh on these bare bones. In 
Chapters 9-23 we will tell you about the most important classes of organic reaction in more detail. 
One of the things organic chemists do, for all sorts of reasons, is to make molecules. And making 
organic molecules means making C-C bonds. In this chapter we are going to look at one of the most 
important ways of making C-C bonds: using organometallics, such as organolithiums and Grignard 
reagents, and carbonyl compounds. We will consider reactions such as these. 



You met these types of reactions in 
Chapter 6: in this chapter we will be 
adding more detail with regard to the 
nature of the organometallic reagents 
and what sort of molecules can be 
made using the reactions. 





new C-C bond 
89% yield 

HO. H 




2. H + , H 2 



new C-C bond 
90% yield 



2. H + , H 2 




"MgCI 



new C-C bond 
80% yield 



2. H + , H 2 




new C-C bond 
75% yield 



The organometallic reagents act as nucleophiles towards the electrophilic carbonyl group, and 
this is the first thing we need to discuss: why are organometallics nucleophilic? We then move on to, 
firstly, how to make organometallics, then to the sort of electrophiles they will react with, and then 
finally to the sort of molecules we can make with them. 

Organometallic compounds contain a carbon-metal bond 

The polarity of a covalent bond between two different elements is determined by electronegativity. 
The more electronegative an element is, the more it attracts the electron density in the bond. So the 



210 



9 ■ Using organometallic reagents to make C-C bonds 



How important are organometallics for making C-C bonds? 



As an example, let's take a molecule known as 
'juvenile hormone'. It is a compound that prevents 
several species of insects from maturing and can be 
used as a means of controlling insect pests. Only very 
small amounts of the naturally occurring compound 
can be isolated, but it can instead be made in the lab 




Cecropia juvenile hormone 



C0 2 Me 



from simple starting materials. At this stage you need 
not worry about how, but we can tell you that, of the 
sixteen C-C bonds in the final product, seven were 
made by reactions of organometallics, many of them 
the sort of reactions we will describe in this chapter. 
This is not an isolated example. As further proof, take 



black bonds made by 
organometallic reactions 



this important enzyme inhibitor, closely related to 
arachidonic acid which you met in Chapter 7. It has 
been made by a succession of C-C bond-forming 
reactions using organometallics: eight of the twenty 
C-C bonds in the product were formed using 
organometallic reactions. 



^C0 2 H 



an enzyme inhibitor 



electronegativities 
2.5 3.5 



C=0 7t bond polarized 
towards oxygen 



nucleophiles 
attack here 



electronegativities 
2.5 1.0 



C— Li a bond polarized 
towards carbon 



MeLi attacks 
electrophiles here 



greater the difference between the electronegativities, the greater the difference between the attrac- 
tion for the bonding electrons, and the more polarized the bond becomes. In the extreme case of 
complete polarization, the covalent bond ceases to exist and is replaced by electrostatic attraction 
between ions of opposite charge. We discussed this in Chapter 4 (p. 000), where we considered the 
extreme cases of bonding in NaF. 

When we discussed (in Chapter 6) the electrophilic nature of carbonyl groups we saw that their 
reactivity is a direct consequence of the polarization of the carbon-oxygen bond towards the more 
electronegative oxygen, making the carbon a site for nucleophilic attack. In organolithium com- 
pounds and Grignard reagents the key bond bond is polarized in the opposite direction — towards 
carbon — making carbon a nucleophilic centre. This is true for most organometallics because, as you 
can see from this edited version of the periodic table, metals (such as Li, Mg, Na, K, Ca, and Al) all 
have lower electronegativity than carbon. 

Pauling electronegativities of selected elements 



H 
2.2 





Li 


Be 






















B 


C 


N 





F 




1.0 


1.6 






2.0 


2.5 


3.04 


3.5 


4.0 




Na 


Mg 


Al 


Si 


P 


s 


CI 




0.9 


1.3 






1.6 


1.9 


2.2 


2.6 


3.2 




K 


Ca 


















Cu 


Zn 








Se 


Br 




0.8 


1.0 


















1.9 


1.7 








2.6 


3.0 





The orbital diagram — the kind you met in Chapter 4 — represents the C-Li bond in methyl- 
lithium in terms of a sum of the atomic orbitals of carbon and lithium. Remember that, the more 



orbital diagram for the C-Li bond of MeLi 



CD 
cz 
CD 



0* MO 



2s \ 






'a MO 



Li- 



these three orbitals are 
involved in C-H bonds 



,-H; ~± ~± ~± 



q To To To 

sp^ sp J sp J sp J 



-« — c* 



lithium lithium-carbor carbon 
atom atom 




Making organometallics 



211 



electronegative an atom is, the lower in energy its atomic orbitals are (p. 000). The filled C-Li 
orbital that arises is closer in energy to the carbon's sp orbital than to the lithium's 2s orbital, so we 
can say that the carbon's sp orbital makes a greater contribution to the C-Li bond and that the 
C-Li bond has a larger coefficient on carbon. Reactions involving the filled orbital will therefore 
take place at C rather than Li. The same arguments hold for the C-Mg bond of Grignard reagents. 



We can also say that, because the carbon's sp orbital makes a greater contribution to the C-Li 
bond, the bond resembles a filled C sp orbital — in other words it resembles a lone pair on carbon. This 



is a useful idea because it allows us to 
think about the way in which methyl- 
lithium reacts — as though it were an 
ionic compound MeTi + — and you 
may sometimes see MeLi or MeMgCl 
represented in mechanisms as Me~. 



organometallic 



R — Li 



reacts as though it were 



carbanion 







R — MgX reacts as though it were 







metal cation 



© 



© 
MgX 



The true structure of organolithiums and Grignard reagents is rather more 
complicated! 

Even though these organometallic compounds are extremely complex aggregates with two, four, six, or more molecules 

reactive with water and oxygen, and have to be handled bonded together, often with solvent molecules. In this book 

under an atmosphere of nitrogen or argon, a number have we shall not be concerned with these details, and it will 

been studied by X-ray crystallography in the solid state and suffice always to represent organometallic compounds as 

by NMR in solution. It turns out that they generally form simple monomeric structures. 



You have already met cyanide (p. 
000), a carbon nucleophile that 
really does have a lone pair on 
carbon. Cyanide's lone pair is 
stabilized by being in a lower- 
energy sp orbital (rather than sp 3 ) 
and by having the electronegative 
nitrogen atom triply bonded to the 
carbon. 



Carbon atoms that carry a 
negative charge, for example 
Me~, are known as carbanions. 



Making organometallics 

How to make Grignard reagents 

Grignard reagents are made by reacting magnesium turnings with alkyl halides in ether solvents to 
form solutions of alkylmagnesium halide. Iodides, bromides, and chlorides can be used, as can both 
aryl and alkyl halides, though they cannot contain any functional groups that would react with the 
Grignard reagent once it is formed. Here are some examples. 



Mg, THF 



Mg, Et 2 



»- MeMgl 




MgCI 



oxidative insertion 



magnesium(O) 



t 
Mg 



magnesium inserts 
into this bond 






Mg 

A 

magnesium(l 



The reaction scheme is easy enough to draw, but what is the 
mechanism? Overall it involves an insertion of magnesium into the 
new carbon-halogen bond. There is also a change in oxidation state 
of the magnesium, from Mg(0) to Mg(II). The reaction is therefore 
known as an oxidative insertion or oxidative addition, and is a 
general process for many metals such as Mg, Li (which we meet 
shortly), Cu, and Zn. 

The mechanism of the reaction is not completely understood 
but a possible (but probably not very accurate) way of writing the 
mechanism is shown here: the one thing that is certain is that the 
first interaction is between the metal and the halogen atom. 



R can be alkyl 
or aryl 

\ / 

R — X 

Mg, Et 2 

R— Mg-X 

alkylmagnesium halide 
(Grignard reagent) 



Diethyl ether (Et 2 0) and THF are 
the most commonly used 
solvents, but you may also meet 
others such as dimethoxyethane 
(DME)and dioxane. 

common ether solvents 



diethyl ether THF 

(tetrahydrofuran) 



^ x OMe 
""0' DME 

dioxane (dimethoxyethane) 




-*- R— Mg-X 



212 



9 ■ Using organometallic reagents to make C-C bonds 





'M 



The reaction takes place not in solution but on the surface of the metal, and how easy it is to make 
a Grignard reagent can depend on the state of the surface — how finely divided the metal is, for exam- 
ple. Magnesium is usually covered by a thin coating of magnesium oxide, and Grignard formation 
generally requires 'initiation' to allow the metal to come into contact with the alkyl halide. Initiation 
can be accomplished by adding a small amount of iodine or 1,2-diiodoethane, or by using ultra- 
sound to dislodge the oxide layer. The ether solvent is essential for Grignard formation because (1) 
ethers (unlike, say, alcohols or dichloromethane) will not react with Grignards and, more impor- 
tantly, (2) only in ethers are Grignard reagents soluble. In Chapter 5 you saw how triethylamine 
forms a complex with the Lewis acid BF 3 , and much the same happens when an ether meets a metal 
ion such as magnesium or lithium: the metals are Lewis-acidic because they have empty orbitals (2p 
in the case of Li and 3p in the case of Mg) that can accept the lone pair of the ether. 



complex between 

Lewis-acidic metal atom 

and lone pairs of THF 



R can be alkyl 
or aryl 



X can be I, Br 
or CI 



\ 



R — X 

Li, THF 

R — Li LiX 

alkylithium plus lithium halide 



How to make organolithium reagents 

Organolithium compounds may be made by a similar oxidative insertion reaction from lithium 
metal and alkyl halides. Each inserting reaction requires two atoms of lithium and generates one 
equivalent of lithium halide salt. As with Grignard formation, there is really very little limit on the 
types of organolithium that can be made this way. 



MeCI 



Li, Et 2 



-*- MeLi + LiCI 



Li, hexane, 
50 °C 




CI 



Li, pentane 



^N 



Li, THF 



CI 



^^L 




+ LiCI 



+ LiBr 



OMe 



+ LiCI 



+ LiBr 



Some Grignard and organolithium reagents are commercially available 

Most chemists (unless they were working on a very large douse these methods). The table lists some of the most 

scale) would not usually make the simpler organolithiums important commercially available organolithiums and 

or Grignard reagents by these methods, but would buy Grignard reagents, 
them in bottles from chemical companies (who, of course, 



Commercially available organometallics 



methyllithium (MeLi) 



methylmagnesium chloride, bromide, and iodide 
(MeMgX) 



n-butyllithium (n-BuLi orjust BuLi) 



sec-butyllithium (sec-BuLi or s-BuLi) 




ethylmagnesium bromide (EtMgBr) 



butylmagnesium chloride (BuMgCI) 



tert-butyllithium (tert-BuLi or t-BuLi) 
phenyllithium (PhLi) 



allylmagnesium chloride and bromide ; 



MgX 



phenylmagnesium chloride and bromide (PhMgCI or 
PhMgBr) 



Organometallics as bases 

Organometallics need to be kept absolutely free of moisture — even moisture in the air will destroy 
them. The reason is that they react very rapidly and highly exothermically with water to produce 



Making organometallics 



213 



alkanes. Anything that can protonate them will do the same thing. If we represent these protonation 
reactions slightly differently, putting the products on the left and the starting materials (represented, 
just for effect, as 'carbanions') on the right, you can see that that they are acid-base equilibria from 
the last chapter. The organometallic acts as a base, and is protonated to form its conjugate acid — 
methane or benzene in these cases. 



© 



Li-rMiT 



methane 



Me — H 



+ Li 



© 



Me — H + H 2 - —r Me Q + H 3 0® 



pK a = 43 



,© 



BrMg-rPl/ 



benzene 
-*- Ph — H 



+ Mg 20 + Br 



Ph — H + H 2 ' —^ 

pK a = 48 



Ph Q + H 3 0® 



The equilibria lie vastly to the left: the pK a values indicate that methane and benzene are extreme- 
ly weak acids and that methyllithium and phenylmagnesium bromide must therefore be extremely 
strong bases. Some of the most important uses of organolithiums — butyllithium, in particular — are 
as bases and, because they are so strong, they will deprotonate almost anything. That makes them 
very useful as reagents for making other organolithiums. 

Making organometallics by deprotonating alkynes 

In Chapter 8 (p. 000) we talked about how hybridization affects acidity. Alkynes, with their C-H 
bonds formed from sp orbitals, are the most acidic of hydrocarbons, with P-K" a s of about 25. They 
can be deprotonated by more basic organometallics such as butyllithium or ethylmagnesium 
bromide. Alkynes are sufficiently acidic to be deprotonated even by nitrogen bases, and another 
common way of deprotonating alkynes is to use NaNH2 (sodium amide), obtained by reacting 
sodium with liquid ammonia. An example of each is shown here: we have chosen to repre- 
sent the alkynyllithium and alkynylmagnesium halide as organometallics and the alkynyl sodium 
as an ionic salt. Propyne and acetylene are gases, and can be bubbled through a solution of the 
base. 



l-hexyne 
pK a ca. 26 



-H + n-Bu — Li 

n-butyllithium 



THF 



-78 °C 



1-hexynyllithium 



-Li + n-Bu — H 

butane 
pK a ca. 50 



Me- 



propyne 



-H + Et — MgBr 



ethylmagnesium 
bromide 



THF 



20 °C 



Me- 



-MgBr + Et — H 



propynylmagnesium bromide ethane 



ethyne (acetylene) 



Na 



© 



NH? 



-78 °C 



Na 



"sodium acetylide" 



NH 3 

ammonia 
pK a ca. 35 



The metal derivatives of alkynes can be added to carbonyl electrophiles as in the following exam- 
ples. The first (we have reminded you of the mechanism for this) is the initial step of an important 
synthesis of the antibiotic, erythronolide A, and the second is the penultimate step of a synthesis of 
the widespread natural product, farnesol. 



214 



9 ■ Using organometallic reagents to make C-C bonds 




Ethynyloestradiol 

The ovulation-inhibiting component of many oral 
contraceptive pills is a compound known as 
ethynyloestradiol, and this compound too is made by an 
alkynyllithium addition to the female sex hormone 



oestrone. A range of similar synthetic analogues of 
hormones containing an ethynyl unit are used in 
contraceptives and in treatments for disorders of the 
hormonal system. 



Me i JDH 




ethynyloestradiol 



Making organometallics by deprotonating aromatic rings: ortholithiation 

Look at the reaction below: in some ways it is quite similar to the ones we have just been discussing. 
Butyllithium deprotonates an sp hybridized carbon atom to give an aryllithium. It works because 
the protons attached to sp carbons are more acidic than protons attached to sp carbons (though 
they are a lot less acidic than alkyne protons). 



OMe 




OMe 



20 X 



+ Bu — Li 




+ Bu — H 



The terms ortho, meta, and para were 
defined on p. 000. 



But there is another factor involved as well. There has to be a functional group containing oxygen 
(sometimes nitrogen) next to the proton to be removed. This functional group 'guides' the butyl- 
lithium, so that it attacks the adjacent protons. It does this by forming a complex with the Lewis- 
acidic lithium atom, much as ether solvents dissolve Grignard reagents by complexing their 
Lewis-acidic metal ions. This mechanism means that it is only the protons ortho to the functional 
group that can be removed, and the reaction is known as an ortholithiation. 



Making organometallics 



215 



complexation between 
oxygen and Lewis- acidic Li 



MeO 




BuLi 




The example below shows an organolithium formed by ortholithiation being used to make a new 
C-C bond. Here it is a nitrogen atom that directs attack of the butyllithium. 
,NMe 2 ^NMe 2 Me 2 N, 




BuLi 




l.Ph' 



2. H + , H 2 




73% yield 



Ortholithiation is useful because the starting material does not need to contain a halogen atom. 
But it is much less general than the other ways we have told you about for making organolithiums, 
because there are rather tight restrictions on what sorts of groups the aromatic ring must carry. 



Fredericamycin 

Fredericamycin is a curious aromatic compound extracted 
in 1981 from the soil bacterium Streptomyces griseus. It 
is a powerful antibiotic and antitumour agent, and its 
structure is shown below. The first time it was made in the 
laboratory, in 1988, the chemists in Boston started their 
synthesis with three consecutive lithiation reactions: two 
are ortholithiations, and the third is slightly different. You 
needn't be concerned about the reagents that react with 
the organolithiums; just look at the lithiation reactions 

sec-BuLi: slightly more 
basic than n-BuLi 



themselves. In each one, an oxygen atom (colour-coded 
green) directs a strongly basic reagent to remove a nearby 
proton (colour-coded black). As it happens, none of the 
steps uses n-BuLi itself, but instead its more reactive 
cousins, sec-BuLi and tert-BuLi (see the table on p. 000). 
The third lithiation step uses a different kind of base, 
made by deprotonating an amine (pK a about 35). The 
yellow proton removed in this third lithiation is more acidic 
because it is next to an aromatic ring (p. 000). 




green oxygens direct RLi to remove yellow protons 



reagent 



tert-BuLi: even more 
basic than sec-BuLi 



NEt 2 



a base made by depronating an amine 

t 

NEt 2 0" 




ortholithiation 



EtO 




OMe 



216 



9 ■ Using organometallic reagents to make C-C bonds 




Br 



Bu — 




Bu — Br 



pK aH = ca. 50 




Br 



<J> 



Halogen-metal exchange 

Deprotonation is not the only 
way to use one simple 
organometallic reagent to gen- 
erate another more useful one. 
Organolithiums can also 
remove halogen atoms from 
alkyl and aryl halides in a reac- 
tion known as halogen-metal 
exchange. Look at this exam- 
ple and you will immediately 
see why. 

The bromine and the lithium simply swap places. As with many of these organometallic process- 
es, the mechanism is not altogether clear, but can be represented as a nucleophilic attack on bromine 
by the butyllithium. But why does the reaction work? The key, again, is pi«C a . The reaction works 
because the organolithium that is formed (phenyllithium, which protonated would give benzene, 
piC a about 43) is less basic (more stable) than the organolithium we started with (BuLi, which proto- 
nated would give butane, pK a about 50). The following reactions are also successful halogen-metal 
exchanges, and in each case the basicity of the organolithium decreases. 

-Br /^ ±\ Br Li 



Bu — Li 



Ph — Li 



can be 

considered 

as 

can be 

considered 

as 



O ,,© 



Bu Li 



Ph© Li® 



pK aH = ca. 43 




n-BuLi 




NR, 



f-BuLi 



NR 2 



+ BuBr 

t-BuLi 



+ f-BuBr 



f-BuLi 



+ f-Bul 



+ t-Bul 



Iodides, bromides, and chlorides can all be used, but the reactions are fastest with iodides and 
bromides. In fact, halogen-metal exchange can be so fast that, at very low temperature (-100 °C and 
below), it is even occasionally possible to use compounds containing functional groups that would 
otherwise react with organolithiums, such as esters and nitro compounds. 



C0 2 Me 



C0 2 Me 





+ BuBr 



Br 



NO; 



n-BuLi 



-100 °C 




NO, 



+ BuBr 



Fenarimol 

Fenarimol is a fungicide that works by inhibiting the 



made by reaction of adiarylketonewith an organolithium 



fungus's biosynthesis of important steroid molecules. It is derived by halogen-metal exchange. 

Ck 



a 



Br 



BuLi 



Fenarimol 




Making organometallics 



217 



► fert-Butyllithium 

Alkyl substituents are slightly electron-donating, so more substituted organolithiums are less stable because the 
carbon atom is forced to carry even more of a negative charge. Instability reaches a peak with tert-butyl lithium, 
which is the most basic of the commonly available organolithium reagents, and so is particularly useful for 
halogen exchange reactions. (It is so unstable that even in solution it will spontaneously catch fire in contact with 
air.) Its importance is enhanced by a subtlety in its reactions that we have not yet mentioned: a problem with 
halogen-metal exchanges is that the two products, an organolithium and an alkyl halide, sometimes react with 
one another in a substitution (Chapter 17) or elimination (Chapter 19) reaction. This problem is overcome 
provided two equivalents of t-BuLi are used. The first takes part in the halogen-metal exchange, while the second 
immediately destroys the t-butyl bromide produced by the exchange, preventing it from reacting with the 
organolithium product. 



Br 



first molecule of r-BuL 
undergoes halogen 
metal exchange 




required organolithium 



inert hydrocarbon 
by-products 



second molecule of 
r-BuLi destroys f-butyl 
halide product 



Do not be concerned about the mechanism at this stage: we will come back to this sort of reaction in Chapter 19. 



Transmetallation 

Organolithiums can be converted to other types of organometallic reagents by transmetallation — 
simply treating with the salt of a less electropositive metal. The more electropositive lithium goes 
into solution as an ionic salt, while the less electropositive metal (magnesium and cerium in these 
examples) takes over the alkyl group. 



MgBr2 



CeCU 



R — MgBr + LiBr 

Grignard 


-* 

dry Et20 or THF 


R — Li 

alkyllithium 


»~ 

dry Et 2 or THF 


R — CeCI 2 + LiCI 

organocerium 



You will see several examples of 
transmetallation with copper salts in 
the next chapter. 



But why bother? Well, the high reactivity — and in particular the basicity — of organolithiums, 
which we have just been extolling, sometimes causes unwanted side-reactions. You saw in Chapter 8 
that protons next to carbonyl groups are moderately acidic (pK a about 20), and because of this 
organolithiums occasionally act as bases towards carbonyl compounds instead of as nucleophiles. 
Organoceriums, for example, are rather less basic, and may give higher yields of the nucleophilic 
addition products than organolithiums or Grignard reagents. 



You met the idea that carbonyl groups, 
and aromatic rings, acidify adjacent 
protons in Chapter 8 (p. 000). 



An instance where transmetallation is needed to produce another 
organometallic, which does act as a base but not as a nucleophile! 



Grignard 



Dialkylzincs are stable, distillable liquids that can be 
made by transmetallating Grignard reagents with zinc 
bromide. They are much less reactive than organolithium 
ororganomagnesium compounds, but they are still rather 
basic and react with water to give zinc hydroxides and 



alkanes. They are used to preserve old books from 
gradual decomposition due to acid in the paper. The 
volatile dialkylzinc penetrates the pages thoroughly, 
where contact with water produces basic hydoxides that 
neutralize the acid, stopping the deterioration. 



dialkylzinc 



R— MgBr 
ZnBr 2 

R — Zn-R 

+ 
H 2 



MgBr 

2 

Zn(0H) 2 + RH 

basic zinc hydroxide 



Acidic protons were a major problem in several syntheses of the anticancer compounds, 
daunorubicin and adriamycin, which start with a nucleophilic addition to a ketone with a pair of 
particularly acidic protons. Organolithium and organomagnesium compounds remove these pro- 



218 



9 ■ Using organometallic reagents to make C-C bonds 



tons rather than add to the carbonyl group, so some Japanese chemists turned to organocerium 
compounds. They made ethynylcerium dichloride (HC=CCeCl2) by deprotonating acetylene, and 
then transmetallating with cerium trichloride. They found that it reacted with the ketone to give an 
85% yield of the alcohol they wanted. 



organolithiums act as bases 
towards this ketone: 



transmetallation gives less 
basic organocerium which 
acts as a nucleophile 



OMe 




Li- 



acidic protons next to 
OMe DOtn carbonyl group and 
aromatic ring shown in 
green 




OMe 



daunorubicin and 
adriamycin 




85% yield 



OMe 



Using organometallics to make organic molecules 

Now that you have met all of the most important ways of making organometallics (summarized here 
as a reminder), we shall move on to consider how to use them to make molecules: what sorts of elec- 
trophiles do they react with and what sorts of products can we expect to get from their reactions? 
Having told you how you can make other organometallics, we shall really be concerned for the rest of 
this chapter only with Grignard reagents and organolithiums. In nearly all of the cases we shall talk 
about, the two classes of organometallics can be used interchangeably. 

• Ways of making organometallics 

• Oxidative insertion of Mg into alkyl halides 

• Oxidative insertion of Li into alkyl halides 

• Deprotonation of alkynes 

• Ortholithiation of functionalized benzene rings 

• Halogen-metal exchange 

• Transmetallation 



Making carboxylic acids from organometallics and carbon dioxide 

Carbon dioxide is a carbonyl com- carbon dioxide 



pound, and it is an electrophile. It 
reacts slowly with water, for exam- 
ple, to form the unstable com- 
pound carbonic acid — you can 
think of this as a hydration reac- 
tion of a carbonyl group. 



n 

o=c=o 






H,0: 



«A4 



carbonic acid 



HO. 



C=0 



HO 



Using organometallic reagents to make organic molecuies 



219 



Grignard reagent 
R — MgBr 



1. C0 2 , Et 2 



2. H 3 + 



=C=!=0 



BrMg 



u car 



AT- 

e 

MgBr 



carboxylic acid 



Carbon dioxide reacts with organo- 
lithiums and Grignard reagents to give car- 
boxylate salts. Protonating the salt 
with acid gives a carboxylic acid with one 
more carbon atom than the starting 
organometallic. The reaction is usually 
done by adding solid CO2 to a solution of 
the organolithium in THF or ether, but it 
can also be done using a stream of dry CO2 
gas. 

The examples below show the three stages of the reaction: (1) forming the organometallic; (2) 
reaction with the electrophile (CO2); and (3) the acidic work-up or quench, which protonates the 
product and destroys any unreacted organometallic left over at the end of the reaction. The three 
stages of the reaction have to be monitored carefully to make sure that each is finished before the 
next is begun — in particular it is absolutely essential that there is no water present during either of 
the first two stages — water must be added only at the end of the reaction, when the organometallic 
has all been consumed by reaction with the electrophile. You may occasionally see schemes written 
out without the quenching step included — but it is nonetheless always needed. 



Br 



m organometallics 

oxidative insertion 
Mg 


>' 


stage 1- formation of 
.the organometallic 

MgBr 

1. C0 2 


stage 
.thee 


2- reaction with 
ectrophile 

COOH 

I 


dry Et 2 


\X\ 


2. H 3 + 


" stage 


3- acid quench 



86% 




Me 



Methicillin synthesis 

Methicillin is an important antiobiotic compound because 
it works even against bacteria that have developed 
resistance to penicillin, whose structure is quite similar. It 
can be made from an acid obtained by reaction of carbon 




OMe 



1. C0 2 



2. HCI, H 2 



dioxide with an organolithium. In this case the 
organolithium is made by an ortholithiation reaction of a 
compound with two oxygen atoms that direct removal of 
the proton in between them. 



OMe 



cc 



several 
more 
steps 



.OMe 



OMe 



OMe 



OMe 




C0 2 H 



Making primary alcohols from organometallics and formaldehyde 

You met formaldehyde, the simplest aldehyde, in Chapter 6, where we discussed the difficuties of 
using it in anhydrous reactions: it is either hydrated or a polymer (paraformaldehyde, (CH 2 0)„) 
and, in order to get pure, dry formaldehyde, it is necessary to heat ('crack') the polymer to 



220 



9 ■ Using organometallic reagents to make C-C bonds 



decompose it. But formaldehyde is a remarkably useful reagent for making primary alcohols, in 
other words, alcohols that have just one carbon substituent attached to the hydroxy-bearing C atom. 
Just as carbon dioxide adds one carbon and makes an acid, fomaldehyde adds one carbon and makes 
an alcohol. 



Primary, secondary, and tertiary are 
defined on p. 000. 



a primary alcohol from formaldehyde 



primary alcohol with one 
additional carbon atom 




CI 



Mg, Et 2 




MgCI 



1. CH 2 

i 

2. H 3 + 




N 0H 



69% 



MgCI 




Ah® 







In the next examples, formaldehyde makes a primary alcohol from two deprotonated 
alkynes. The second reaction here (for which we have shown organolithium formation, reaction, 
and quench simply as a series of three consecutive reagents) forms one of the last steps of the 
synthesis of Cecropia juvenile hormone whose structure you met right at the beginning of the 
chapter. 



Ph- 



n-BuLi 



-*- Ph- 



1. (CH 2 0) n 

-Li »~ Ph- 

2. H 3 + 



\ 



91% 



OH 




2. (CH 2 0)„ 

3. H 3 + 



i Something to bear in mind with all organometallic additions to carbonyl 
compounds is that the addition takes the oxidation level down one. In other 
words, if you start with an aldehyde, you end up with an alcohol. More 
specifically, 



• Additions to CO2 give carboxylic acids 

• Additions to formaldehyde (CH2O) give primary alcohols 



VI 

I 

it^oh 



• Additions to other aldehydes (RCHO) give secondary alcohols ^/\ 

R 3 

• Additions to ketones give tertiary alcohols r2 \J 



OH 



Secondary and tertiary alcohols: which organometallic, which aldehyde, which 
ketone? 

Aldehydes and ketones react with Grignard or organolithium reagents to form secondary and ter- 
tiary alcohols, respectively, and some examples are shown with the general schemes here. 



Using organometallic reagents to make organic molecules 



221 



secondary alcohols from aldehydes 

aldehyde secondary alcohol 

OH 

1. R 2 MgBr 

-^u 2. H,0 + *~ Ri 



tertiary alcohols from ketones 

ketone tertiary alcohol 



R 1 H 



R 1 ^R 2 



VI 

X 



1. R 3 MgBr 



\ 



OH 



BrMg 



&% 



H .0 



r-* 



e 



R 1 "H 



y/ 



R 1 



R 1 ^R 2 2. H 3 + 



^P - 



R^ R 



R 3 .0 



r s 



R 1 "R" 



© 
MgBr 



R 1 "R 



two examples: 



) MgCI 



two examples 




2. H 3 + 




54% 





1. Me — MgCI 
2 H 3 + 



To make any secondary alcohol, however, there is often a choice of two possible routes, depend- 
ing on which part of the molecule you choose to make the organometallic and which part you choose 
to make the aldehyde. For example, the first example here shows the synthesis of a secondary alcohol 
from isopropylmagnesium chloride and acetaldehyde. But it is equally possible to make this same 
secondary alcohol from isobutyraldehyde and methyllithium or a methylmagnesium halide. 



81% 



acetaldehyde 



Me-^^0 



) MgCI 

2. H 3 + 

54% yield 




isobutyraldehyde 
1. Me — MgCI ^^ 



2. H 3 + 



69% yield 



^o 



Indeed, back in 1912, when this alcohol was first described in detail, the chemists who made it chose 
to start with acetaldehyde, while in 1983, when it was needed as a starting material for a synthesis, it was 
made from isobutyraldehyde. Which way is better? The 1983 chemists probably chose the isobu- 
tyraldehyde route because it gave a better yield. But, if you were making a secondary alcohol for the first 
time, you might just have to try both in the lab and see which one gave a better yield. Or you might be 
more concerned about which uses the cheaper, or more readily available, starting materials — this was 
probably behind the choice of methylmagnesium chloride and the unsaturated aldehyde in the second 



Flexible alcohol synthesis 

As an illustration of the flexibility available in 
making secondary alcohols, one synthesis of 
bongkrekic acid, a highly toxic compound that 
inhibits transport across certain membranes in the 

alcohols needed for the synthesis of bongkrekic acid 



R 3 Si 



cell, required both of these (very similar) alcohols. 
The chemists making the compound at Harvard 
University chose to make each alcohol from quite 
different starting materials: an unsaturated 



R 3 Si 



aldehyde and an alkyne-containing organolithium in 
the first instance, and an alkyne-containing 
aldehyde and vinyl magnesium bromide in the 
second. 




2. H + , H 2 




222 



9 ■ Using organometallic reagents to make C-C bonds 



example. Both can be bought commercially, while the alternative route to this secondary alcohol would 
require a vinyllithium or vinylmagnesium bromide reagent that would have to be made from a vinyl 
halide, which is itself not commercially available, along with difficult-to-dry acetaldehyde. 



Note we have dropped the aqueous 
quench step from these schemes to 
avoid cluttering them. 



u PrO, ° 



R 1 ' 

oxidation 



commercially 
commercially available 

available 



1. Me — MgCI 



2. H 3 + 




ally commercially available 
but hard to dry 



Me 



2. H 3 + 



With tertiary alcohols, there is even more choice. The last example in the box is a step in a synthe- 
sis of the natural product, nerolidol. But the chemists in Paris who made this tertiary alcohol could 
in principle have chosen any of these three routes, 
three routes to a tertiary alcohol 




Only the reagents in orange are commercially available, but, as it happens, the green Grignard 
reagent can be made from an alkyl bromide, which is itself commercially available, making the route 
on the left the most reasonable. 

Now, do not be dismayed! We are not expecting you to remember a chemical catalogue and to 
know which compounds you can buy and which you can't. All we want you to appreciate at this stage 
is that there are usually two or three ways of making any given secondary or tertiary alcohol, and you 
should be able to suggest alternative combinations of aldehyde or ketone and Grignard reagent that 
will give the same product. You are not expected to be able to assess the relative merits of the differ- 
ent possible routes to a compound. That is a topic we leave for a much later chapter on retrosynthet- 
ic analysis, Chapter 30. 

Ketones by oxidation of secondary alcohols 

Tertiary alcohols can be made from ketones, and secondary ones from aldehydes, but we should now 
show you that ketones can be made from secondary alcohols by an oxidation reaction. There are lots 
of possible reagents, but a common one is an acidic solution of chromium trioxide. We will look in 
much more detail at oxidation later, when we will discuss the mechanism of the reaction, but for 
now take it from us that secondary alcohols give ketones on treatment with CrC>3. Note that you 
can't oxidize tertiary alcohols (without breaking a C-C bond). The link between secondary alcohols 
and ketones means that the ketones needed for making tertiary alcohols can themselves ultimately be 
made by addition of organometallics to carbonyl compounds. Here, for example, is a sequence of 
reactions leading to a compound needed to make the drug viprostol. 



^ 



MgBr 



"CHO 



2. H,0 + 



n-pentyl 




Cr0 3 



1. R 2 



j 



2. H 3 + 



OH 



A closer look at some mechanisms 



223 



BrMg 



^J£ 



A closer look at some mechanisms 

We finish this chapter with some brief words about the mechanism of the addition of 
organometallics to carbonyl compounds. The problem with this reaction is that no-one really knows 
precisely what happens during the addition reaction. We know what the organic products are 
because we can isolate them and look at them using NMR and other spectroscopic techniques. But 
what happens to the metal atoms during the reaction? 

You will have noticed that we always write the addition reaction with the metal atom just falling 
off the organometallic as it reacts, and then appearing near to the anionic oxygen atom of the prod- 
uct. In other words, we have not been specific about what the metal atom is actually doing during the 
addition; in fact, we have been deliberately vague so as not to imply anything that may not be true. 
But there is one thing that is certain about this process, and before we discuss it we need to remind 
you of something we talked about in Chapter 6: the effect of acid on the addition of nucleophiles to 
carbonyl groups. We said that acid tends to catalyse addition reactions by protonating the carbonyl 
group, making it positively charged and therefore more electrophilic. 

Now, of course, in our organometallic addition reactions we have no acid (H + ) present, because 
that would destroy the organometallic reagent. But we do have Lewis-acidic metal atoms — Li or 
Mg — and these can play exactly the same role. They can coordinate to the carbonyl's oxygen atom, 
giving the carbonyl group positive charge and therefore making it more electrophilic. In one possible 
version of the mechanism, a four-centred mechanism allows coordination of the magnesium to the 
oxygen while the nucleophilic carbon atom attacks the carbonyl group. The product ends up with a 
(covalent) Mg-O bond, but this is just another way of writing RCF MgBr + . 



R 

X. 







© 

MgBr 



Lewis acids were introduced in Chapter 
5. 



Br. 



~Mg f \ 



BrMg. 








which can also 
be written as 




© 
MgBr 



clotted Bond 
indicates new 
bond forming 



dotted bond 
indicates old ■ 
bond breaking 



Bk 






Mg 

I 

I 

Im- 



possible transition 

state for 
Grignard addition 



The four-centred mechanism is quite hard to visualize just 
with curly arrows: what they are saying is that the O-Mg inter- 
action is forming at the same time as the new C-C bond, and that 
simultaneously the old C-Mg bond and C=0 % bond are break- 
ing. A neat way of representing all of this is to draw what we might 
see if we took a snapshot of the reaction halfway through, using 
dotted lines to represent the partially formed or partially broken 
bonds. It would look something like this, and such a snapshot is 
known as the transition state for the reaction. 
An alternative possibility is that two molecules of the Grignard reagent are involved, and that the 
transition state is a six-membered ring. We are telling you all this not because we want to confuse 
you but because we want to be honest: there is genuine uncertainty about the mechanism, and this 
arises because, while it is easy to determine the products of a reaction using spectroscopy, it is much 
harder to determine mechanisms. 



The term 'transition state' has, in 
fact, a more precise definition, 
which we will introduce in Chapter 
13. 



We shall devote two chapters entirely 
to mechanism and how it is studied: 
Chapters 13 and 41. 




>^>MgBr 



Mg 
0"' '~Br 



.MgBr 



six-centred transition state 



,MgR 




MgBr 2 



But, for one type of Grignard reagent, it is certain that the addition proceeds through a six-mem- 
bered ring. Here is a reaction between an allylic Grignard reagent and a ketone. The product is a ter- 
tiary alcohol, but perhaps not the tertiary alcohol you would expect. The Grignard reagent appears to 



224 



9 ■ Using organometallic reagents to make C-C bonds 



have attached itself via the wrong carbon atom. We can explain this by a six-membered transition 
state, but one involving only one molecule of Grignard reagent. 



Barbier was Victor Grignard's PhD 
supervisor. 



Mg, Et 2 




dimer 



Grignard reagent reacts 
through this carbon 




81% yield 



curly-arrow representation of the mechanism 




J 





six-centred transition state 



Allylic Grignard reagents are unusual for more than one reason, and it turns out that they are, in 
fact, quite hard to make in good yield from allyl halides. The problem is that the allyl halide is highly 
reactive towards the Grignard reagent as it forms, and a major by-product tends to be a dimer. The 
way round this problem is to make the Grignard reagent actually in the presence of the carbonyl 
compound. This method works in a number of cases, not just with allylic Grignards, and is often 
called the Barbier method. 

For example, it is a straightforward matter to make these three alcohols, provided the allylic 
halide, aldehyde, and magnesium are all mixed together in one flask. The Grignard reagent forms, 
and immediately reacts with the aldehyde, before it has a chance to dimerize. In the second example, 
notice again that the allylic Grignard reagent must have reacted through a six-membered transition 
state because the allyl system has 'turned around' in the product. 

OH OH CIMg s 



70% yield 




80% yield 





The last reaction above leads us nicely into the next chapter where we will look at an alternative 
way for such unsaturated aldehydes to react — by conjugate addition. 



Problems 



1. Propose mechanisms for the first four reactions in the chapter 
OH 

»- 

2. H + , H 2 





1. 




MgBr HO, 



2. H + , H 2 




HO. H 




1.. 



HO. H 



s MgCI 



2. H + , H 2 




Problems 



225 



2. When this reaction is carried out with allyl bromide labelled as 
shown with C, the label is found equally distributed between the 
ends of the allyl system in the product. Explain how this is 
possible. How would you detect the C distribution in the 
product? 



OH 



13 c 



Br 1. Mg, THF 

*• 

2. PhCHO 

3. H + , H 2 



Ph 



3. What products would be formed in these reactions? 

1. EtMgBr 
Ph ^= H »~ A 



^ 



2. Ph 2 C0 
1. Mg, THF 



Br 



-*- B 



Br 




CI 



3. H + , H 2 

1. BuLi 
»~ C 

2. C0 2 

3. H + , H 2 



4. Suggest alternative routes to fena- 
rimol — that is, different routes from 
the one shown in the chapter. 




5. The synthesis of the gastric antisecreto- 
ry drug rioprostil requires this alcohol. 

(a) Suggest possible syntheses starting from ketones and 
organometallics and (b) suggest possible syntheses of the ketones 
in part (a) from aldehydes and organometallics (don't forget 
about CrC>3 oxidation!). 

6. Suggest two syntheses of the bee 
pheromone heptan-2-one. 

heptan-2-one 

7. How could you prepare these compounds using ortho- 
lithiation procedures? 

P P 

MeO> 




xx!r 




8. Why is it possible to make the lithium derivative A by Br/Li 
exchange, but not the lithium derivative B? 



BuLi 




THF 
BuLi 
THF 




9. Comment on the selectivity (that is, say what else might have 
happened and why it didn't) shown in this Grignard addition 
reaction used in the manufacture of an antihistamine drug. 




10. The antispasmodic drug biperidin is made by the Grignard 
addition reaction shown here. What is the structure of the drug? 
Do not be put off by the apparent complexity of the com- 
pounds — the chemistry is the same as that you have seen in this 
chapter. How would you suggest that the drug procyclidine 

should be madei. 

Br 1. Mg, Et 2 

biperidin 



// 






11. Though heterocyclic compounds, such as the nitrogen ring 
system in this question, are introduced rather later in this book, 
use your knowledge of Grignard chemistry to draw a mechanism 
for what happens here. It is important that you prove to yourself 
that you can draw mechanisms for reactions on compounds that 
you have never met before. 



MeO 




226 



9 ■ Using organometallic reagents to make C-C bonds 



12. What product would be formed in this reaction between a 
chloro compound and a seven-membered ring ketone? 

CU ^ !■ M g. THF 




N. 



Me 




Conjugate addition 



10 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• Reactions of C=0 groups ch6 & ch9 


• How conjugation affects reactivity 


• Conjugate addition in other 


• Conjugation ch7 


• What happens to a C=0 group when it 


electrophilic alkenes ch23 




is conjugated with a C=C bond 


• Conjugate addition with further types 




• How the C=C double bond becomes 


of nucleophiles ch29 




electrophilic, and can be attacked by 


• Alkenes that are not conjugated with 




nucleophiles 


C=0ch20 




• Why some sorts of nucleophiles attack 






C=C while others still attack the C=0 






group 





Conjugation changes the reactivity of carbonyl groups 

To start this chapter, here are four reactions of the same ketone. For each product, the principal 
absorptions in the IR spectrum are listed. The pair of reactions on the left should come as no surprise 
to you: nucleophilic addition of cyanide or a Grignard reagent to the ketone produces a product with 
no C=0 peak near 1700 cm , but instead an O-H peak at 3600 cm . The 2250 cm peak is C=N; 
C=C is at 1650 cm 



Me 



Me 



-1 



NaCN, HCN 
5-10 °C 



NC. OH 



Me 



Me 



NaCN, HCN 
80 °C 



-*~ A 



IR: 3600 (broad), 2250, 1650 
no absorption near 1700 




1. BuMgBr 

2. H 2 



Bu OH 




Me 



IR: 3600 (broad), 1640 
no absorption near 1700 



Me 




IR: 2250, 1715 
no absorption at 3600 

1. BuMgBr, 1% CuCI 

2. H 2 

*- 



IR: 1710 
no absorption at 3600 



If you need to review IR spectroscopy, 
turn back to Chapter 3. Chapter 6 dealt 
with addition of CN~ to carbonyl 
compounds, and Chapter 9 with the 
addition of Grignard reagents. 



But what about the reactions on the right? Both products A and B have kept their carbonyl group 
(IR peak at 1710 cm~ ) but have lost the C=C. Yet A, at least, is definitely an addition product Me 
because it contains a C^N peak at 2200 cm - . 

Well, the identities of A and B are revealed here: they are the products of addition, not to 
the carbonyl group, but to the C=C bond. This type of reaction is called conjugate addition, and is 
what this chapter is all about. The chapter will also how explain how such small differences in 
reaction conditions (temperature, or the presence of CuCI) manage to change the outcome 
completely. 



direct addition to the C=0 group 



NC 



Me 




NC 






© 



NC OH 



Me 



Me 



CN 



Bu 



228 



10 ■ Conjugate addition 



conjugate addition to the C=C double bond 



Jit"* 



c? 



Me 



6 








CN 



Me 



CN 



The a and (3 refer to the distance 
of the double bond from the C=0 
group: the a carbon is the one 
next to C=0 (notthe carbonyl 
carbon itself), the p carbon is one 
further down the chain, and so on. 




cc,P-unsaturated ketone 




l,Y-unsaturated ketone 



Conjugate addition to the C=C double bond follows a similar course to direct addition to the 
C=0 group, and the mechanisms for both are shown here. Both mechanisms have two steps: addi- 
tion, followed by protonation. Conjugate additions only occur to C=C double bonds next to C=0 
groups. They don't occur to C=C bonds that aren't immediately adjacent to C=0 (see the box on p. 
000 for an example). 

Compounds with double bonds adjacent to a C=0 group are known as a,P-unsaturated carbonyl 

compounds. Many a,(3-unsaturated carbonyl compounds have trivial names, and some are shown 

here. Some classes of a,[3-unsaturated carbonyl compounds also have names such as 'enone' or 

'enal', made up of 'ene' (for the double bond) + 'one' (for ketone) or 'ene' + 'aP (for aldehyde). 

an a,|3-unsaturated aldehyde an a,p-unsaturated ketone an a,p-unsaturated acid an a,p-unsaturated ester 
(an enal) (an enone) 



HO 



propenal 
(trivial name = acrolein) 



but-3-en-2-one 

(trivial name = 

methyl vinyl ketone) 



propenoic acid 

(trivial name = 

acrylic acid) 



EtO 

ethyl propenoate 
(trivial name = 
ethyl acrylate) 



A range of nucleophiles will undergo conjugate additions with a,(3-unsaturated carbonyl com- 
pounds, and six examples are shown below. Note the range of nucleophiles, and also the range of car- 
bonyl compounds: esters, aldehydes, acids, and ketones. 

types of nucleophile which 
undergo conjugate addition 



cyanide 



KCN 



HCN 



OMe 



CN 



OMe 



Et 2 NH + 



OEt 



100 °c 



Et 2 N 



OEt 



alcohols 



MeOH + 




Ca(OH) 2 



OMe 



thiols 



bromide 



MeSH + 



HBr + 



OH 



NaOH 



MeS' ^" 

O 



Br 




OH 



chloride 



HCI + %. 



CI 



Polarization is detectable spectroscopically 



229 



The reason that OC,P-unsaturated carbonyl compounds react differently is conjugation, the phe- 
nomenon we discussed in Chapter 7. There we introduced you to the idea that bringing two % sys- 
tems (two C=C bonds, for example, or a C=C bond and a C=0 bond) close together leads to a 
stabilizing interaction. It also leads to modified reactivity, beacuse the 7t bonds no longer react as 
independent functional groups but as a single, conjugated system. 



Termite self-defence and the reactivity of alkenes 

Soldier termites of the species Schedorhinotermes lamanianus defend their 
nests by producing this compound, which is very effective at taking part in 
conjugate addition reactions with thiols (RSH). This makes it highly toxic, since 
many important biochemicals carry SH groups. The worker termites of the same 
species — who build the nests — need to be able to avoid being caught in the 
crossfire, so they are equipped with an enzyme that allows them to reduce 
compound 1 to compound 2. This still has a double bond, but the double bond 
is completely unreactive towards nucleophiles because it is not conjugated with 
a carbonyl group. The workers escape unharmed. 



compound 1 




reacts with 
nucleophiles 



compound 2 



Alkenes conjugated with carbonyl groups are polarized 

You haven't met many reactions of alkenes yet: detailed discussion will have to wait till Chapter 20. 
But we did indicate in Chapter 5 that they react with electrophiles. Here is the example from 
p. 000: in the addition of HBr to isobutene the alkene acts as a nucleophile and H-Br as the 
electrophile. 

H-LBr 

2 



Me. 



Me 



(^Br 
Me\ H 



CHo 



Me H 



Me 
Me 



>— <J-H 



C=C double bond acts 
as a nucleophile 



curly arrows indicate 
derealization of electrons 



Me 




O ~ 



£° 



Me 



© 



This is quite different to the reactivity of a C=C 
double bond conjugated with a carbonyl group, 
which, as you have just seen, reacts with nucleophiles 
such as cyanide, amines, and alcohols. The conjugated 
system is different from the sum of the isolated parts, 
with the C=0 group profoundly affecting the reactivi- 
ty of the C=C double bond. To show why, we can use 
curly arrows to indicate derealization of the 71 electrons over the four atoms in the conjugated sys- 
tem. Both representations are extremes, and the true structure lies somewhere in between, but the 
polarized structure indicates why the conjugated C=C bond is electrophilic. 



true electron distribution lies somewhere 
in between these extremes 



> Conjugation makes alkenes electrophilic 

• Isolated C=C double bonds are 
nucleophilic 



C=C double bonds conjugated 
with carbonyl groups are 




electrophilic 



Co 



© 




Nu e 



Polarization is detectable spectroscopically 

IR spectroscopy provides us with evidence for polarization in C=C bonds conjugated to C=0 bonds. 
An unconjugated ketone C=0 absorbs at 1715 cm~ while an unconjugated alkene C=C absorbs 



You maybe asking yourself why 
we can't show the derealization 
by moving the electrons the other 
way, like this. 



Mo 



M 



,AJ> 



Think about electronegativities: 
is much more electronegative 
than C, so it is quite happy to 
accept electrons, but here we 
have taken electrons away, 
leaving it with only six electrons. 
This structure therefore cannot 
represent what happens to the 
electrons in the conjugated 
system. 



230 



10 ■ Conjugate addition 



(usually rather weakly) at about 1650 crrT . Bringing these two groups into conjugation in an 
a,(3-unsaturated carbonyl compound leads to two peaks at 1675 and 1615 cm , respectively, both 
quite strong. The lowering of the frequency of both peaks is consistent with a weakening of both 7t 
bonds (notice that the polarized structure has only single bonds where the C=0 and C=C double 
bonds were). The increase in the intensity of the C=C absorption is consistent with polarization 
brought about by conjugation with C=0: a conjugated C=C bond has a significantly larger dipole 
moment than its unconjugated cousins. 

The polarization of the C=C bond is also evident in the C NMR spectrum, with the signal for 
the sp carbon atom furthest from the carbonyl group moving downfield relative to an unconjugated 
alkene to about 140 p. p.m., and the signal for the other double bond carbon atom staying at about 
120 p.p.m. 




143 p.p.m. 



compared with 




132 p.p.m. 



124 p.p.m. 



119 p.p.m. 



electrons must move from 
HOMO of nucleophile 



MeO 



°o£ 



MeO 




In acrolein, the HOMO is in fact 
not the highest filled jt orbital you 
see here, but the lone pairs on 
oxygen. This is not important 
here, though, because we are 
only considering acrolein as an 
electrophile, so we are only 
interested in its LUMO. 



Molecular orbitals control conjugate additions 

We have spectroscopic evidence that a conjugated C=C bond is polarized, and we can explain this 
with curly arrows, but the actual bond-forming step must involve movement of electrons from the 
HOMO of the nucleophile to the LUMO of the unsaturated carbonyl compound. The example in the 
margin has methoxide (MeO - ) as the nucleophile. 



^/^ 



butadiene 



^^r 



acrolein 



But what does this LUMO 
look like? It will certainly be 
more complicated than the Jt* 
LUMO of a simple carbonyl 
group. The nearest thing you 
have met so far (in Chapter 7) 
are the orbitals of butadiene 
(C=C conjugated with C=C), 
which we can compare with 
the a,(3-unsaturated aldehyde 
acrolein (C=C conjugated with 
C=0). The orbitals in the 71 sys- 
tems of butadiene and acrolein 
are shown here. They are dif- 
ferent because acrolein's orbi- 
tals are perturbed (distorted) 
by the oxygen atom (Chapter 
4). You need not be concerned 
with exactly how the sizes of 
the orbitals are worked out, but 
for the moment just concen- 
trate on the shape of the 
LUMO, the orbital that will 
accept electrons when a 
nucleophile attacks. 

In the LUMO, the largest coefficient is on the (3 carbon of the a,(3-unsaturated system, shown 
with an asterisk. And it is here, therefore, that nucleophiles attack. In the reaction you have just 
seen, the HOMO is the methoxide oxygen's lone pair, so this will be the key orbital interaction 





Ammonia and amines undergo conjugate addition 



231 



that gives rise to the new bond. The second largest coefficient is on the C=0 carbon atom, so it's not 
surprising that some nucleophiles attack here as well — remember the example right at the begin- 
ning of the chapter where you saw cyanide attacking either the double bond or the carbonyl group 
depending on the conditions of the reaction. We shall next look at some conjugate additions with 
alcohols and amines as nucleophiles, before reconsidering the question of where the nucleophile 
attacks. 



Me- 



G 



HOMO = sp 3 on 



LUMO 




Me- 



new a bond 




Ammonia and amines undergo conjugate addition 

Amines are good nucleophiles for conjugate addition reactions, and give products that we can term 
P-amino carbonyl compounds (the new amino group is P to the carbonyl group). Dimethylamine is 
a gas at room temperature, and this reaction has to be carried out in a sealed system to give the 
ketone product. 




H 
\ 
Me — N 

Me 



Me 



Me 2 NH 

1 

50 °C, 1 h 





50% yield 



This is the first conjugate addition mechanism we have shown you that involves a neutral nucle- 
ophile: as the nitrogen adds it becomes positively charged and therefore needs to lose a proton. We 
can use this proton to protonate the negatively charged part of the molecule as you have seen hap- 
pening before. This proton-transfer step can alternatively be carried out by a base: in this addition of 
butylamine to an 0C,f3-unsaturated ester (ethyl acrylate), the added base (EtCT) deprotonates the 
nitrogen atom once the amine has added. Only a catalytic amount is needed, because it is regenerat- 
ed in the step that follows. 

n-BuNH, 



OEt 



KOEt, EtOH 
30 °C 




99% yield 



BuNH 




OEt 



0> Et 



Bu' 



OEt 




Q 3 



OEt 



Ammonia itself, the simplest amine, is very volatile (it is a gas at room temperature, but a very 
water-soluble one, and bottles of 'ammonia' are actually a concentrated aqueous solution of ammo- 
nia), and the high temperatures required for conjugate addition to this unsaturated carboxylic acid 
can only be achieved in a sealed reaction vessel. 



232 



10 ■ Conjugate addition 



Tertiary amines can't give 
conjugate addition products 
because they have no proton to 
lose. 



H H 

I 
OH 

hydroxylamine 



NH, 



MeS 



NH 3 , H 2 



MeS 



OH 



OH 



150 °C in a sealed tube 



64% yield 

Amines are bases as well as nucleophiles, and in this reaction the first step must be deprotonation 
of the carboxylic acid: it's the ammonium carboxylate that undergoes the addition reaction. You 
would not expect a negatively charged carboxylate to be a very good electrophile, and this may well 
be why ammonia needs 150 °C to react. 

C N " 3 If NH 3 



MeS 



°3 



H 



MeS 







-*- MeS 



OH 



The (3-amino carbonyl product of conjugate addition of an amine is still an amine and, provided 
it has a primary or secondary amino group, it can do a second conjugate addition. For example, 
methylamine adds successively to two molecules of this unsaturated ester. 



OMe 



MeNH, 



MeHN 




OMe OMe OMe 

77% yield 
Two successive conjugate additions can even happen in the same molecule. In the next example, 
hydroxylamine is the nucleophile. Hydroxylamine is both an amine and an alcohol, but it always 
reacts at nitrogen because nitrogen (being less electronegative than oxygen) has a higher-energy 
(more reactive) lone pair. Here it reacts with a cyclic dienone to produce a bicyclic ketone, which we 
have also drawn in a perspective view to give a better idea of its shape. 



o 



NH 2 0H 



MeOH 



NOH )=0 

77% yield 



can be 
drawn as 





B h OH 




3|k> 







HOHN 



redrawn as u „ ou 

H H^l 

HO — \ f 




The reaction sequence consists 
of two conjugate addition reac- 
tions. The first is intermolecular, 
and gives the intermediate enone. 
The second conjugate addition is 
intramolecular, and turns the 
molecule into a bicyclic structure. 
Again, the most important steps 
are the C-N bond-forming reac- 
tions, but there are also several 
proton transfers that have to 
occur. We have shown a base 'B:' 
carrying out these proton trans- 
fers: this might be a molecule of 
hydroxylamine, or it might be a 
molecule of the solvent, methanol. 
These details do not matter. 



Conjugate addition of alcohols can be catalysed by acid or base 



233 



NaOMe 



Conjugate addition of alcohols can be catalysed by acid or base 

Alcohols undergo conjugate addition only very slowly in the absence of a catalyst: they are not such 
good nucleophiles as amines for the very reason we have just mentioned in connection with the reac- 
tivity of hydroxylamine — oxygen is more electronegative than nitrogen, and so its lone pairs are of 
lower energy and are therefore less reactive. Alkoxide anions are, however, much more nucleophilic. 
You saw methoxide attacking the orbitals of acrolein above: the reaction in the margin goes at less 
than5°C. 

The alkoxide doesn't have to be made first, though, because alcohols dissolved in basic solution 
are at least partly deprotonated to give alkoxide anions. How much alkoxide is present depends on 
the pH of the solution and therefore the pK a of the base (Chapter 8), but even a tiny amount is 
acceptable because once this has added it will be replaced by more alkoxide in acid-base equilibrium 
with the alcohol. In this example, allyl alcohol adds to pent-2-enal, catalysed by sodium hydroxide in 
the presence of a buffer. 



/s 



CHO 



CHO 



OMe 




small amount of 
alkoxide produced 



In Chapter 6 we discussed the role of 
base and acid catalysts in the direct 
addition of alcohols to carbonyl 
compounds to form hemiacetals. The 
reasoning — that base makes 
nucleophiles more nucleophilic and 
acid makes carbonyl groups more 
electrophilic — is the same here. 



Only a catalytic amount of base is required as the deprotonation of ROH (which can be water or 
allyl alcohol) in the last step regenerates more alkoxide or hydroxide. It does not matter that sodium 
hydroxide (pK a n 15.7) is not basic enough to deprotonate an alcohol (p-K" a 16-17) completely, since 
only a small concentration of the reactive alkoxide is necessary for the reaction to proceed. 

We can also make rings using alkoxide nucleophiles, and in this example the phenol (hydroxy- 
benzene) is deprotonated by the sodium methoxide base to give a phenoxide anion. Intramolecular 
attack on the conjugated ketone gives the cyclic product in excellent yield. In this case, the methoxide 
(piC a H about 16) will deprotonate the phenol (pK a about 10) completely, and competitive attack by 
MeO~ acting as a nucleophile is not a problem as intramolecular reactions are usually faster than 
their intermolecular equivalents. 



NaOMe 

93% yield 




H— OMe 



c-ia 



There are some important exceptions 
to this depending on the size of ring 
being formed, and some of these are 
described in Chapter 42. 



OMe 



234 



10 ■ Conjugate addition 



Acid catalysts promote conjugate addition of alcohols to (X,P-unsaturated carbonyl compounds 
by protonating the carbonyl group and making the conjugated system more electrophilic. Methanol 
adds to this ketone exceptionally well, for example, in the presence of an acid catalyst known as 
'Dowex 50'. This is an acidic resin — just about as acidic as sulfuric acid in fact, but completely in- 
soluble, and therefore very easy to remove from the product at the end of the reaction by filtration. 

MeOH 
Dowex 50 



MeOH 




Once the methanol has added to the protonated enone, all that remains is to reorganize the pro- 
tons in the molecule to give the product. This takes a few steps, but don't be put off by their com- 
plexity — as we've said before, the important step is the first one — the conjugate addition. 



Conjugate addition or direct addition to the carbonyl group? 

We have shown you several examples of conjugate additions using various nucleophiles and a,(3- 
unsaturated carbonyl compounds, but we haven't yet addressed one important question. When do 
nucleophiles do conjugate addition (also called '1,4-addition') and when do they add directly to the 
carbonyl group ('1,2-addition')? Several factors are involved — they are summarized here, and we 
will spend the next section of this chapter discussing them in turn. 



conjugate addition to C=C 
(also called "1,4-addition") 



direct addition to C=0 
(also called "1,2-addition" 



c° 



p<>v 



Nu 1 



e 



Nu 1 



0- 



Z> 



The way that nucleophiles react depends on: 

• the conditions of the reaction 

• the nature of the OC,p-unsaturated carbonyl compound 

• the type of nucleophile 



Reaction conditions 

The very first conjugate addition reaction in this chapter depended on the conditions of the reaction. 
Treating an enone with cyanide and an acid catalyst at low temperature gives a cyanohydrin by direct 
attack at C=0, while heating the reaction mixture leads to conjugate addition. What is going 
on? 



NaCN, HCN, 
5-10 °C 



NC. OH 



NaCN, HCN, 
80 °C 



CN 



cyanohydrin 
(direct addition to carbonyl) 



conjugate addition product 



Conjugate addition or direct addition to the carbonyl group? 



235 



We'll consider the low-temperature reaction first. As you know from Chapter 6, it is quite normal 
for cyanide to react with a ketone under these conditions to form a cyanohydrin. Direct addition to 
the carbonyl group turns out to be faster than conjugate addition, so we end up with the cyanohydrin. 



" CN° ?\ CN NC 0H 



slow but fast but 

conjugate addition product irreversible reversible cyanohydrin 

thermodynamic product: kinetic product: 

more stable forms faster 

Now, you also know from Chapter 6 that cyanohydrin formation is reversible. Even if the equilib- 
rium for cyanohydrin formation lies well over to the side of the products, at equilibrium there will 
still be a small amount of starting enone remaining. Most of the time, this enone will react to form 
more cyanohydrin and, as it does, some cyanohydrin will decompose back to enone plus cyanide — 
such is the nature of a dynamic equilibrium. But every now and then — at a much slower rate — the 
starting enone will undergo a conjugate addition with the cyanide. Now we have a different situa- 
tion: conjugate addition is essentially an irreversible reaction, so once a molecule of enone has been 
converted to conjugate addition product, its fate is sealed: it cannot go back to enone again. Very 
slowly, therefore, the amount of conjugate addition product in the mixture will build up. In order 
for the enone-cyanohydrin equilibrium to be maintained, any enone that is converted to conjugate 
addition product will have to be replaced by reversion of cyanohydrin to enone plus cyanide. Even at 
room temperature, we can therefore expect the cyanohydrin to be converted bit by bit to conjugate 
addition product. This may take a very long time, but reaction rates are faster at higher temperatures, 
so at 80 °C this process does not take long at all and, after a few hours, the cyanohydrin has all been 
converted to conjugate addition product. 

The contrast between the two products is this: cyanohydrin is formed faster than the conjugate 
addition product, but the conjugate addition product is the more stable compound. 

Typically, kinetic control involves lower temperatures and shorter reaction times, which ensures 
that only the fastest reaction has the chance to occur. And, typically, thermodynamic control 
involves higher temperatures and long reaction times to ensure that even the slower reactions have a 
chance to occur, and all the material is converted to the most stable compound. 

• Kinetic and thermodynamic control 

• The product that forms faster is called the kinetic product 

• The product that is the more stable is called the thermodynamic product 
Similarly, 

• Conditions that give rise to the kinetic product are called kinetic control 

• Conditions that give rise to the thermodynamic product are called thermo- 
dynamic control 

Why is direct addition faster than conjugate addition? Well, although the carbon atom (3 to the 
C=0 group carries some positive charge, the carbon atom of the carbonyl group carries more, and so 
electrostatic attraction for the charged nucleophiles will encourage it to attack the carbonyl group 
directly rather than undergo conjugate addition. 

attack is possible at but electrostatic attraction to 

either site C=0 is greater 

t 
8+ 8+y 




LUMO 



236 



10 ■ Conjugate addition 



And why is the conjugate addition product the more stable? In the conjugate addition product, 

we gain a C-C bond, losing a C=C 71 bond, but keeping the C=0 7t bond. With direct addition, we 

still gain a C-C bond, but we lose the C=0 Tt bond and keep the C=C Tt bond. C=0 Tt bonds are 

stronger than C=C 71 bonds, so the conjugate addition product is the more stable. 

lose C=0 Ji bond gain C-C a bond 

369 kJ mol" 1 fl 9 

gain C-C bond . .---"' 




~CN 



lose C=C Tt bond 
280 kJ mor 1 " 
We will return to kinetic and thermodynamic control in Chapter 13, where we will analyse the 
rates and energies involved a little more rigorously, but for now here is an example where conjugate 
addition is ensured by thermodynamic control. Note the temperature! 




HCN, KCN 
160 °C 




75% yield 



CN 



CI 



a,(5-unsaturated acyl chloride 



enal 



OR 

a,|3-unsaturated ester 


NR 2 

a,p-unsaturated amide 



BuLi, -70 
H 2 



C to +20 




BuLi, 
H 2 



70 °C to +20 °C 



Bu 



Structural factors 

Not all additions to carbonyl groups are reversible: additions of organometallics, for example, 
are certainly not. In such cases, the site of nucleophilic attack is determined simply by reactivity: 
the more reactive the carbonyl group, the more direct addition to C=0 will result. The most reactive 
carbonyl groups, as you will see in Chapter 12, are those that are not conjugated with O or N (as 
they are in esters and amides), and particularly reactive are acyl chlorides and aldehydes. In 
general, the proportion of direct addition to the carbonyl group follows the reactivity sequence in the 
margin. 

Compare the way butyllithium 
adds to this 0C,(3-unsaturated alde- 
hyde and OC,P-unsaturated amide. 
Both additions are irreversible, and 
BuLi attacks the reactive carbonyl 
group of the aldehyde, but prefers 

conjugate addition to the less reac- ^NMe 2 "~NMe 2 

tive amide. Similarly, ammonia 
reacts with this acyl chloride to give 
an amide product that derives (for 
details see Chapter 12) from direct 
addition to the carbonyl group, 
while with the ester it undergoes 
conjugate addition to give an ^OMe H 2 N'' ^OMe 

amine. 

Sodium borohydride is a nucleophile that you have seen reducing simple aldehydes and ketones 
to alcohols, and it usually reacts with CC,p%unsaturated aldehydes in a similar way, giving alcohols by 
direct addition to the carbonyl group. 

NaBH 4 , EtOH OH 



NH 3 



CI 



NH, 



NH 3 



,-CHO 



97% yield 




NaBH„, EtOH 




99% yield 



Quite common with ketones, though, is the outcome on the right. The borohydride has reduced 



Conjugate addition or direct addition to the carbonyl group? 



237 



not only the carbonyl group but the double bond as well. In fact, it's the double bond that's reduced 
first in a conjugate addition, followed by addition to the carbonyl group. 




conjugate 
addition 




a second 

addition direct 

to C=0 



9^y 



o> k 





MeO 




NaBH 4 , MeOH 



MeO 




Ph 



Ph 



hJc^h 

/\ 
H H 

For esters and other less reactive carbonyl com- 
pounds conjugate addition is the only reaction that 
occurs. 

Steric hindrance also has a role to play: the more 
substituents there are at the (3 carbon, the less likely a 
nucleophile is to attack there. Nonetheless, there are plenty of examples where nucleophiles undergo 
conjugate addition even to highly substituted carbon atoms. 

The nature of the nucleophile: hard and soft 

Among the best nucleophiles of all at doing conjugate addition are thiols, the sulfur analogues of 
alcohols. In this example, the nucleophile is thiophenol (phenol with the O replaced by S). 
Remarkably, no acid or base catalyst is needed (as it was with the alcohol additions), and the product 
is obtained in 94% yield under quite mild reaction conditions. 

3H PhSH, 25 °C, 

R— SH 





5h 



SPh 



94% 



a thiol 



thiophenol 

Why are thiols such good nucleophiles for conjugate additions? Well, to explain this, and why 
they are much less good at direct addition to the C=0 group, we need to remind you of some ideas 
we introduced in Chapter 5. There we said that the attraction between nucleophiles and electrophiles 
is governed by two related interactions — electrostatic attraction between positive and negative 
charges and orbital overlap between the HOMO of the nucleophile and the LUMO of the elec- 
trophile. Successful reactions usually result from a combination of both, but sometimes reactivity 
can be dominated by one or the other. The dominant factor, be it electrostatic or orbital control, 
depends on the nucleophile and electrophile involved. Nucleophiles containing small, electonegative 
atoms (such as O or CI) tend to react under predominantly electrostatic control, while nuclophiles 
containing larger atoms (including the sulfur of thiols, but also P, I, and Se) are predominantly sub- 
ject to control by orbital overlap. The terms 'hard' and 'soft' have been coined to describe these two 
types of reagents. Hard nucleophiles are typically from the early rows of the periodic table and have 
higher charge density, while soft nucleophiles are from the later rows of the periodic table — they are 
either uncharged or have larger atoms with higher-energy, more diffuse orbitals. 

Table 10. 1 divides some nucleophiles into the two categories (plus some that lie in between) — but 
don't try to learn it! Rather, 
convince yourself that the 
properties of each one justify 
its location in the table. Most of 
these nucleophiles you have 
not yet seen in action, and the 
most important ones at this 
stage are indicated in bold 
type. 



Table 10.1 Hard and soft nucleophiles 




Hard nucleophiles Borderline 


Soft nucleophiles 


r, OH", RO", SO^, CI", N5, CN" 


r, RS", RSe~, S 2 ~ 


H 2 0, ROH, ROR', RCOR', RNH 2 , RR'NH, 


RSH, RSR', R 3 P 


NH 3 , RMgBr, RLi Br" 


alkenes, aromatic rings 



This reaction, and how to control 
reduction of C=0 and C=C, will be 
discussed in more detail in Chapter 24. 



The concept of steric hindrance was 
introduced in Chapter 6. 




R t » 



C T3 

o co 



238 



10 ■ Conjugate addition 



Not only can nucleophiles be classified as hard or soft, but electrophiles can too. For example, H + 
is a very hard electrophile because it is small and charged, while Br 2 is a soft electrophile: its orbitals 
are diffuse and it is uncharged. You saw Br 2 reacting with an alkene earlier in the chapter, and we 
explained in Chapter 5 that this reaction happens solely because of orbital interactions: no charges 
are involved. The carbon atom of a carbonyl group is also a hard electrophile because it carries a par- 
tial positive charge due to polarization of the C=0 bond. What is important to us is that, in general, 
hard nucleophiles prefer to react with hard electrophiles, and soft nucleophiles with soft elec- 
trophiles. So, for example, water (a hard nucleophile) reacts with aldehydes (hard electrophiles) to 
form hydrates in a reaction largely controlled by electrostatic attraction. On the other hand, water 
does not react with bromine (a soft electrophile). Yet bromine reacts with alkenes while water does 
not. Now this is only a very general principle, and you will find plenty of examples where hard reacts 
with soft and soft with hard. Nonetheless it is a useful concept, which we shall come back to later in 
the book. 

• Hard/soft reactivity 

• Reactions of hard species are dominated by charges and electrostatic effects 

• Reactions of soft species are dominated by orbital effects 

• Hard nucleophiles tend to react well with hard electrophiles 

• Soft nucleophiles tend to react well with soft electrophiles 

What has all this to do with the conjugate addition of thiols? Well, an a,(3-unsaturated carbonyl 
compound is unusual in that it has two electrophilic sites, one of which is hard and one of which is 
soft. The carbonyl group has a high partial charge on the carbonyl carbon and will tend to react with 
hard nucleophiles, such as organolithium and Grignard reagents, that have a high partial charge on 
the nucleophilic carbon atom. Conversely,the (3 carbon of the a,(3-unsaturated carbonyl system does 
not have a high partial positive charge but is the site of the largest coefficient in the LUMO. This 
makes the P carbon a soft electrophile and likely to react well with soft nucleophiles such as thiols. 

• Hard/soft — direct/conjugate addition 

• Hard nucleophiles tend to react at the carbonyl carbon (hard) of an enone 

• Soft nucleophiles tend to react at the p-carbon (soft) of an enone and lead to 
conjugate addition 



Anticancer drugs that work by conjugate addition of thiols 




Drugs to combat cancer act on a range of 
biochemical pathways, but most commonly on 
processes that cancerous eel Is need to use to 
proliferate rapidly. One class attacks DNA 
polymerase, an enzyme needed to make the copy of 
DNA that has to be provided for each new cell. 
Helenalin and vernolepin are two such drugs, and if 
you look closely at their structure you should be 
able to spot two a,p-unsaturated carbonyl groups in 



each. Biochemistry is just chemistry in very small 
flasks called cells, and the reaction between DNA 
polymerase and these drugs is simply a conjugate 
addition reaction between a thiol (the SH group of 
one of the enzyme's cysteine residues) and the 
unsaturated carbonyl groups. The reaction is 
irreversible, and shuts down completely the 
function of the enzyme. 



vernolepin 



Enzyme 




Copper(l) salts have a remarkable effect on organometallic reagents 



239 



Copper(I) salts have a remarkable effect on organometallic 
reagents 

Grignard reagents add directly to the carbonyl group of a,(3-unsaturated aldehydes and ketones to 
give allylic alcohols: you have seen several examples of this, and you can now explain it by saying that 
the hard Grignard reagent prefers to attack the harder C=0 rather than the softer C=C electrophilic 
centre. Here is a further example — the addition of MeMgl to a cyclic ketone to give an allylic alcohol, 
plus, as it happens, some of a diene that arises from this alcohol by loss of water (dehydration). Below 
this example is the same reaction to which a very small amount (just 0.01 equivalents, that is, 1%) of 
copper(I) chloride has been added. The effect of the copper is dramatic: it makes the Grignard 
reagent undergo conjugate addition, with only a trace of the diene. 




Me 




MeMgBr 
Et 2 



Me 




43% 



48% 




MeMgBr 

CuCI (0.01 eq) 

Et 2 



Me 



Me — 7k Jr — Me 

Me Me 

83% 




Me 



Organocopper reagents undergo conjugate addition 

The copper works by transmetallating the Grignard reagent to give an organocopper reagent. 
Organocoppers are softer than Grignard reagents, and add in a conjugate fashion to the softer C=C 
double bond. Once the organocopper has added, the copper salt is available to transmetallate some 
more Grignard, and only a catalytic amount is required. 



Me — MgBr 




Me 

conjugate Me 

addition of *\ J 

organocopper 



Organocoppers are softer than 
Grignard reagents because 
copper is less electropositive 
than magnesium, sotheC-Cu 
bond is less polarized than the 
C-Mg bond, giving the carbon 
atom less of a partial negative 
charge. Electronegativities: Mg, 
1.3; Cu, 1.9. 



Me 



Me 



er / \ 

■*- / j^0 S MgBr® 

Me 1 



H,0 



copper(l) recycled: only a catalytic quantity is requi 



The organocopper is shown here as 'Me-Cu' because its pre- 
cise structure is not known. But there are other organocopper 
reagents that also undergo conjugate addition and that are much 
better understood. The simplest result from the reaction of two 
equivalents of organolithium with one equivalent of a copper (I) 
salt such as CuBr in ether or THF solvent at low temperature. The 
lithium cuprates (R^CuLi) that are formed are not stable and 
must be used immediately. 



Me 



lithium 
cuprate 
reagent 




We discussed 
transmetallation in 
Chapter 9. 



Me 



2 x R — Li 



CuBr 



Et 2 
-78 °C 



^Cu Li® 



+ LiBr 



As with the organolithiums that we 
introduced in Chapter 9, the exact 
structure of these reagents is 
more complex than we imply here: 
they are probably tetramers (four 
molecules of R 2 CuLi bound 
together), but for simplicity we will 
draw them as monomers. 



240 



10 ■ Conjugate addition 



The addition of lithium cuprates to OC,fi-unsaturated ketones turns out to be much better if 
trimethylsilyl chloride is added to the reaction — we will explain what this does shortly, but for the 
moment here are two examples of lithium cuprate additions. 

1. Ph 2 CuLi, Me 3 SiCI OMe 1. Bu 2 CuLi, Me 3 SiCI 

2H + ,H 2 _ 1 ^ _ 2. H + ,H 2 



OMe 



-*- Ph 



^/ 



CHO 



„CH0 



75% yield 



80% yield 



The silicon works by reacting with the negatively charged intermediate in the conjugate addition 
reaction to give a product that decomposes to the carbonyl compound when water is added at the 
end of the reaction. Here is a possible mechanism for a reaction between Bu2CuLi and an a,(3-unsat- 
urated ketone in the presence of Me3SiCl. The first step is familiar to you, but the second is a new 
reaction. Even so, following what we said in Chapter 5, it should not surprise you: the oxygen is 
clearly the nucleophile and the silicon the electrophile, and a new bond forms from O to Si as indi- 
cated by the arrow. The silicon-containing product is called a silyl enol ether, and we will come back 
to these compounds and their chemistry in more detail in later chapters. 




CuLi ^k Me' ^ 



y< Li@ 




_SiMe 3 




BiT Bu 



Bu 

99% yield 



Conclusion 

We end with a summary of the factors controlling the two modes of addition to a,(3-unsaturated car- 
bonyl compounds, and by noting that conjugate addition will be back again — in Chapters 23 (where 
we consider electrophilic alkenes conjugated with groups other than C=0) and 29 (where the nucle- 
ophiles will be of a different class known as enolates). 



• Summary 








Conjugate addition favoured by 


Direct addition to C=0 favoured by 


Reaction conditions 


• thermodynamic control: high 


• kinetic control: low temperatures, 


(for reversible additions): 


temperatures, long reaction times 


short reaction times 


Structure of a,(l-unsaturated 


• unreactive C=0 group (amide, ester) 


• reactive C=0 group group (aldehyde, 


compound: 




acyl chloride) 




• unhindered (i carbon 


• hindered fl carbon 


Type of nucleophile: 


• soft nucleophiles 


• hard nucleophiles 


Organometallic: 


• organocoppers or catalytic Cu(I) 


• organolithiums, Grignard reagents 



Problems 



241 



Problems 

1. Draw mechanisms for this reaction and explain why this par- 
ticular product is formed. 

H 2 S, NaOAc 
,C0 2 Me *JVIe0 2 C^ 



*^ 1 



,C0 2 Me 



H 2 0, EtOH 



2. Which of the two routes shown here would actually lead to the 
product? Why? 

1. EtMgBr, 2. HCI 



HO. 



OR 



-*~ CI 



1. HCI, 2. EtMgBr 

3. Suggest reasons for the different outcomes of the following 
reactions (your answer must, of course, include a mechanism for 
each reaction). 



OH 



LiAIH 4 



R 2 NH 



H,H 2 Aj^L 



MeC0 2 



4. Addition of dimethylamine to the unsaturated ester A could 
give either product B or C. Draw mechanisms for both reactions 
and show how you would distinguish them spectroscopically. 


Me 2 NH 
Me 2 N' >" ^OMe -«- 





OMe 




Me 2 NH 



5. Suggest mechanisms for the following reactions. 

NaOAc 

^^N0 2 



NMe 2 



HOAc 



A- 



,N0 2 



OMe 



OMe 



MeHN 




6. Predict the product of these reactions 
.OMe 



Me0 2 C 




i-PrMgCI, CuSPh 



Me 




7. Two routes are proposed for the preparation of this amino 
alcohol. Which do you think is more likely to succeed and why? 

1. ^.CHO 

*~ 





2. LiAIH 4 



8. How would you prepare these compounds by conjugate addi- 
tion? 




Me 2 N. 



.CN 



9. How might this compound be made using a conjugate addi- 
tion as one of the steps? You might find it helpful to consider the 
preparation of tertiary alcohols as decribed in Chapter 9 and also 
to refer back to Problem 1 in this chapter. 



HO 



N 
I 
Me 



OH 



10. When we discussed reduction of cyclopentenone to cyclo- 
pentanol, we suggested that conjugate addition of borohydride 
must occur before direct addition of borohydride; in other words, 
this scheme must be followed. 




NaBH, 




NaBH, 




cyclopentenone 



intermediate 
not isolated 



cyclopentnol 



What is the alternative scheme? Why is the scheme shown 
above definitely correct? 



242 



10 ■ Conjugate addition 



11. Suggest a mechanism for this reaction. Why does conjugate 
addition occur rather than direct addition? 




OSiMe 



Ph,P 



Me,SiCI 




Why is the product shown as a cation? If it is indeed a salt, 
what is the anion? 



12. How, by choice of reagent, would you make this reaction give 
the direct addition product (route A)? How would you make it 
give the conjugate addition product (route B)? 

HO. / P 

route A 



route B 




Proton nuclear magnetic resonance 



11 



Connections 



Building on: 

• X-ray crystallography, mass 

spectrometry, 13 C NMR and infrared 
spectroscopy ch3 



Arriving at: 

• Proton (or 1 H) NMR spectroscopy 

• How 1 H NMR compares with 13 C NMR 

• How 'coupling' in 1 H NMR provides 
most of the information needed to find 
the structure of an unknown molecule 



Looking forward to: 

Using 1 H NMR with other 
spectroscopic methods to solve 
structures rapidly chl5 

Using 1 H NMR to investigate the 
detailed shape (stereochemistry) of 
molecules ch32 

1 H NMR spectroscopy is referred to in 
most chapters of the book as it is the 
most important tool for determining 
structure; you must understand this 
chapter before reading further 



The differences between carbon and proton NMR 

We used C NMR in Chapter 3 as part of a three-pronged attack on the problem of determining 
molecular structure. Important though these three prongs are, we were forced to confess at the end 
of Chapter 3 that we had delayed the most important technique of all — proton ( H) NMR — until a 
later chapter because it is more complicated than C NMR. This is that delayed chapter and we 
must now tackle those complications. We hope you will see H NMR for the beautiful and powerful 
technique that it surely is. The difficulties are worth mastering for this is the chemist's primary 
weapon in the battle to solve structures. 

Proton NMR differs from C NMR in a number of ways. 



Three prongs: 13 C NMR; infrared 
spectroscopy; mass spectrometry. 

► 

1 H NMR and proton NMR are 
interchangeable terms. Chemists 
often use 'proton' to mean not 
only H + but also the nucleus of a 
hydrogen atom forming part of a 
molecule. This is how it will be 
used in this chapter. 



H is the major isotope of hydrogen (99.985% natural abundance), 
while C is only a minor isotope (1.1%) 

H NMR is quantitative: the area under the peak tells us the number of 
hydrogen nuclei, while C NMR may give strong or weak peaks from 
the same number of C nuclei 

Protons interact magnetically ('couple') to reveal the connectivity of the 
structure, while C is too rare for coupling between C nuclei to be seen 

H NMR shifts give a more reliable indication of the local chemistry 
than that given by C spectra 



► An instance where Hi NMR was not useful 

In Chapter 3 you met methoxatin. Proton NMR has little to tell us 

about its structure as it has so few protons (it is C 14 H 6 N20 8 ). 

Carbon NMR and eventually an X-ray crystal structure gave the 

answer. There are four OH and NH 

protons (best seen by IR) and only 

two C-H protons. The latter protons 

are the kind that proton NMR 

reveals best. Fortunately, most 

compounds have lots more than 

this. 




We shall examine each of these points in detail and build up a full understanding of proton NMR 
spectra. The other spectra remain important, of course. 
Proton NMR spectra 



are recorded in the same 
way as 13 C NMR spectra: 
radio waves are used to 
study the energy level 
differences of nuclei, but 
this time they are H and 
not C nuclei. Hydrogen 
nuclei have a nuclear spin 



applied 

magnetic 

field B„ 



nucleus aligned 
against applied 
magnetic field 



© 



nucleus aligned 

with applied 

magnetic field 




^ — 



higher 
energy 

level 



lower 

energy 

level 



* 



energy 



The number of energy levels 
available to a nucleus of spin / is 
2/+1. 



244 



11 ■ Proton nuclear magnetic resonance 



of a half and so have two energy levels: they can be aligned either with or against the applied magnetic 
field. 

The spectra look much the same: the scale runs from right to left and the zero point is given by the 
same reference compound though it is the proton resonance of Me4Si rather than the carbon reso- 
nance that defines the zero point. You will notice at once that the scale is much smaller, ranging over 
only about 10 p. p.m. instead of the 200 p. p.m. needed for carbon. This is because the variation in the 
chemical shift is a measure of the shielding of the nucleus by the electrons around it. There 
is inevitably less change possible in the distribution of two electrons around a hydrogen nucleus 
than in that of the eight valence electrons around a carbon nucleus. Here is a simple H NMR 
spectrum. 

^H NMR speclmm of acetic acid 



This 10 p. p.m. scale is not the 
same as any part of the 13 C NMR 
spectrum. It is at a different 
frequency altogether. 



"JL 



i 



— r 

10 



5 



ft {p. p.m.) 



It is not enough simply to measure the 
relative heights of the peaks because, 
as here, some peaks might be broader 
than others. Hence the area under the 
peak is measured. 



Integration tells us the number of hydrogen atoms in each 
peak 

The chemical shift of the twelve hydrogen atoms of the four identical methyl groups in Me4Si 
is defined as zero. The methyl group in the acid is next to the carbonyl group and so slightly de- 
shielded at about 8 2.0 p.p.m. and the acidic proton itself is very deshielded at 5 11.2 p.p.m. The 
same factor that makes this proton acidic — the O-H bond is polarized towards oxygen — also makes 
it resonate at low field. So far things are much the same as in carbon NMR. Now for a difference. 
Notice that the ratio of the peak heights in this spectrum was about 3:1 and that that is also the 
ratio of the number of protons. In fact, it's not the peak height but the area under the peaks that is 
exactly proportional to the number of protons. Proton spectra are normally integrated, that is, the 
area under the peaks is computed and recorded as a line with steps corresponding to the area, like 
this. 



^C 



0.75 cm 



2.25 cm 






I ■ 



— r 

10 



i — ' — ■ — ' — ■ — r 
5 6 (p.p.m.) ° 



Simply measuring the height of the steps with a ruler gives you the ratio of the numbers of protons 
represented by each peak. Knowing the atomic composition from the mass spectrum, we also know 
the distribution of protons of various kinds. Here the heights are 0.75 and 2.25 cm, a ratio of about 
1:3. The compound is C2H4O2 so, since there are 4 H atoms altogether, the peaks must contain 1 x H 
and 3 x H, respectively. 

In the spectrum of 1,4-dimethoxybenzene, there are just two signals in the ratio of 3:2. This time 
the compound is CgH 10 O2 so the true ratio must be 6:4. Assigning the spectrum requires the same 
attention to symmetry as in the case of C spectra. 



Regions of the proton NMR spectrum 



245 



H 
\ 



H 



H 3 C0 — \ — 0CH 3 

/ \ 

H H 



200 



solvent 



„JL 



t3 C NMR 



i 51 ■■ 



1 0i I 



5G 



M 

HjCO — C C — OCHj 

H H 



so I vent J 



- 



U NVR 



-.M 



In this next example it is easy to assign the spectrum simply by measuring the steps in the integral. 
There are two identical methyl groups (CMe2) having 6 Hs, one methyl group by itself having 3 Hs, 
the OH proton (1 H), the CH2 group next to the OH (2 Hs), and finally the CH2CH2 group between 
the oxygen atoms in the ring (4 Hs). 



p.p.m. 






"0 



H3C CH3 



solvent 1 



r 



1 1 1 1 1 1 1 1 r 1 

10 5 p.p.m. 

Proton NMR spectra are generally recorded in solution in deuterochloroform (CDCI3) — that is, 
chloroform with the H replaced by H. The proportionality of the size of the peak to the number of 
protons tells you why: if you ran a spectrum in CHCI3, you would see a vast peak for all the solvent 
Hs because there would be much more solvent than the compound you wanted to look at. Using 
CDCI3 cuts out all extraneous protons. 

Regions of the proton NMR spectrum 

The integration gives useful — indeed essential — information, but it is much more important to 
understand the reasons for the exact chemical shift of the different types of proton. In the last 
example you can see one marked similarity to carbon spectra: protons on saturated carbon atoms 
next to oxygen are shifted downfield to larger 8 values (here 3.3 and 3.9 p.p.m.). The other regions of 
the proton NMR spectrum are also quite similar in general outline to those of C spectra. Here they 



246 



11 ■ Proton nuclear magnetic resonance 



regions of the proton NMR spectrum 



Me 4 Si 



protons on 


protons on 


protons on 


saturated 


saturated 


unsaturated 


unsaturated 


unsaturated 


CH 3 


CH 3 


carbons 


carbons: 


carbons: 


CH 2 


CH 2 


next to 


benzene, 


alkenes 


CH 


CH 


oxygen: 


aromatic 




next to 


not next to 


aldehydes 


hydrocarbons 




oxygen 


oxygen 



10.5 



8.5 



6.5 



4.5 



3.0 5 (p. p.m.) 



0.0 



These regions hold for protons attached to C: protons attached to O or N can come almost any- 
where on the spectrum. Even for C-H signals, the regions are approximate and overlap quite a lot. 
You should use the chart as a basic guide, but you will need a more detailed understanding of proton 
chemical shifts than you did for C chemical shifts. To achieve this understanding, we now need to 
examine each class of proton in more detail and examine the reasons for particular shifts. It is impor- 
tant that you grasp these reasons. An alternative is to learn all the chemical shifts off by heart (not 
recommended). 



In this chapteryou will see a lot of 
numbers — chemical shifts and 
differences in chemical shifts. We 
need these to show that the ideas 
behind 1 H NMR are securely 
based in fact. You do not need to 
learn these numbers. 
Comprehensive tables can be 
found at the end of Chapter 15, 
which we hope you will find useful 
for reference while you are solving 
problems. Again, do not attempt 
to learn the numbers! 



The second two compounds, 
dichloromethane CH2CI2 and 
chloroform CHCI 3 , are commonly 
used as solvents and their shifts 
will become familiarto you if you 
look at a lot of spectra. 



You have seen 5 used as a 
symbol for chemical shift. Now 
that we have two sorts of 
chemical shift— in the 13 C NMR 
spectrum and in the H NMR 
spectrum — we need to be able to 
distinguish them. 8h means 
chemical shift in the ^-H NMR 
spectrum, and 8 C chemical shift 
in the 13 C NMR spectrum. 



Protons on saturated carbon atoms 

Chemical shifts are related to the electronegativity of substituents 

We shall start with protons on 



Table 11.1 Effects of electronegativity 



Atom 

Li 

Si 
N 

F 



Electronegativity 

1.0 

1.9 
3.0 
3.4 
4.0 



Compound 

CH3-IJ 

CH 3 -SiMe 3 
CH3-NH2 
CH3-OH 
CH 3 -F 



L H NMR shift, p.p.m. 

-1.94 

0.0 
2.41 
3.50 
4.27 



saturated carbon atoms. If you 
study Table 11.1 you will see 
that the protons in a methyl 
group are shifted more 
and more as the atom attached 
to them gets more electro- 
negative. 

When we are dealing with 
simple atoms as substituents, 
these effects are straightfor- 
ward and more or less additive. If we go on 
adding electronegative chlorine atoms to a 
carbon atom, electron density is progres- 
sively removed from it and the carbon 
nucleus and the hydrogen atoms attached 
to it are progressively deshielded. 

Proton chemical shifts tell us about chemistry 

The truth is that shifts and electronegativity are not perfectly correlated. The key property is indeed 
electron withdrawal but it is the electron-withdrawing power of the whole substituent in comparison 
with the carbon and hydrogen atoms in the CH skeleton that matters. Methyl groups joined to the 
same element, say, nitrogen, may have very different shifts if the substituent is an amino group 
(CH3-NH2 has 5h for the CH3 group = 2.41 p.p.m.) or a nitro group (CH3-NO2 has 5h 4.33 
p.p.m.). A nitro group is much more electron-withdrawing than an amino group. 

What we need is a quick guide rather than some detailed correlations, and the simplest is this: all 
functional groups except very electron-withdrawing ones shift methyl groups from 1 p.p.m. (where 
you find them if they are not attached to a functional group) downfield to about 2 p.p.m. Very elec- 
tron-withdrawing groups shift methyl groups to about 3 p.p.m. 



^■H NMR shift, p.p.m. 
13 C NMR shift, p.p.m. 



CH3CI 

3.06 

24.9 



CH2CI2 

5.30 
54.0 



CHCI3 

7.27 

77.2 



Protons on saturated carbon atoms 



247 



• 


Approximate chemical shifts for methyl groups 




No electron-withdrawing 


Less electron-withdrawing 


More electron-withdrawing 




functional groups 


functional groups X 


functional groups X 


Me at about 1 p. p.m. 


MeX at about 2 p. p.m. 


MeX at about 3 p. p.m. 






(i.e. add 1 p. p.m.) 


(i.e. add 2 p. p.m.) 




aromatic rings, alkenes, 


carbonyl groups: acids (CO2H), 


oxygen-based groups: 




alkynes 


esters (C0 2 R), ketones (COR), 
nitriles (CN) 


ethers (OR), esters (OCOR) 






amines (NHR) 


amides (NHCOR) 






sulfides (SR) 


sulfones (S0 2 R) 









Me 
Me 



Me 



CI 



Rather than trying to fit these data to some atomic property, even such a useful one as electroneg- 
ativity, we should rather see these shifts as a useful measure of the electron-withdrawing power of the 
group in question. The NMR spectra are telling us about the chemistry. Among the largest shifts pos- 
sible for a methyl group is that caused by the nitro group, 3.43 p.p.m., at least twice the size of the 
shift for a carbonyl group. This gives us our first hint of some important chemistry: one nitro group 
is worth two carbonyl groups when you need electron withdrawal. You have already seen that elec- 
tron withdrawal and acidity are related (Chapter 8) and in later chapters you will see that we can cor- 
relate the anion- stabilizing power of groups like carbonyl, nitro, and sulfone with proton NMR. 

Methyl groups give us information about the structure of molecules 

It sounds rather unlikely that the humble methyl |y| e ci 

group could tell us much that is important about 

molecular structure — but just you wait. We shall look 

at four simple compounds and their NMR spectra — 

just the methyl groups, that is. The first two are the 

acid chlorides on the right. 

The first compound shows just one methyl signal containing 9 Hs at 5h 1.10 p.p.m.. This tells us 
two things. All the protons in each methyl group are the same; and all three methyl groups in the ter- 
tiary butyl (f-butyl, or Me3C-) group are the same. This is because rotation about C-C single bonds, 
both about the CH3-C bond and about the (CH 3 ) 3 C-C bond, is fast. Though at any one instant the 
hydrogen atoms in one methyl group, or the methyl groups in the f-butyl group, may differ, on aver- 
age they are the same. The time-averaging process is fast rotation about a bond. The second com- 
pound shows two 3H signals, one at 1.99 and one at 2.17 p.p.m. Now rotation is slow — indeed the 
C=C double bond does not rotate at all and so the two methyl groups are different. One is on the same 
side of the alkene as (or l cis to') the -COC1 group while the other is on the opposite side (or 'trans). 



Me 







Me 



MeS 




MeS 

Me 




The second pair of compounds contain the CHO 
group. One is a simple aldehyde, the other an amide 
of formic acid: it is DMF, dimethylformamide. The 
first has two sorts of methyl group: a 3H signal at 5h 
1.81 p.p.m. for the SMe group and a 6H signal for the 

CMe 2 group. The two methyl groups in the 6H signal are the same, again because of fast rotation 
about a C-C bond. 

The second compound also has two methyl signals, at 2.89 and 2.98 p.p.m., each 3H, and these are 
the two methyl groups on nitrogen. Restricted rotation about the N-CO bond must be making the 
two Me groups different. You will remember from Chapter 7 (p. 000) that the N-CO amide bond 
has considerable double bond character because of conjugation: the lone pair electrons on nitrogen 
are delocalized into the carbonyl group. 



Rotation about single bonds is 
generally very fast (you are about 
to meet an exception); rotation 
about double bonds is generally 
very, very slow (it just doesn't 
happen). This was discussed in 
Chapter 7. 



Me 

I 



H 



Me 
®lL .0© 



Me' 



248 



11 ■ Proton nuclear magnetic resonance 



Chemical shifts of CH2 groups 

Shifts of the same order of magnitude occur for protons on CH2 groups and the proton on CH 
groups, but with the added complication that CH2 groups have two other substituents and CH 
groups three. A CH2 (methylene) group resonates at 1.3 p. p.m., about 0.4 p. p.m. further downfield 
than a comparable CH3 group (0.9 p.p.m.), and a CH (methine) group resonates at 1.7 p.p.m., 
another 0.4 p.p.m. downfield. Replacing each hydrogen atom in the CH 3 group by a carbon atom 
causes a small downfield shift as carbon is slightly more electronegative (C 2.5 p.p.m.; H 2.2 p.p.m.) 
than hydrogen and therefore shields less effectively. 

• Chemical shifts of protons in CH, CH2, and CH3 groups with no nearby 
electron-withdrawing groups 

CH group CH 2 group CH 3 group 

0.4 p.p.m. downfield 0.4 p.p.m. downfield 



8(CH 2 ) -3.0 p.p.m. 

H H 




C0 2 H 

NH 2 

phenylalanine 



You'll meet this reaction in the next 
chapter, and we shall discuss 
protection and protecting groups in 
Chapter 24. For the moment, just be 
concerned with the structure of the 
product. 



1.7 p.p.m. 



1.3 p.p.m. 



0.9 p.p.m. 



The benzyl group (PI1CH2-) is very important in organic chemistry. It occurs naturally in the 
amino acid phenylalanine, which you met in Chapter 2. Phenylalanine has its CH2 signal at 3.0 
p.p.m. and is moved downfield from 1.3 p.p.m. mostly by the benzene ring. 

Amino acids are often protected as the 'Cbz' derivatives (Carboxybenzyl) by reaction with an acid 
chloride. 

Here is a simple example together with the NMR spectrum of the product. Now the CH2 group 
has gone further downfield to 5.1 p.p.m. as it is next to both oxygen and phenyl. 



H ? N 



X 



C0 2 H 



Ph' 



r-^ci 



Ph' 



XX 



H 



C0 2 H 



amino acid 



"Cbz chloride" 
(benzyl chloroformate) 



"Cbz protected" amino acid 



COO 






Ph 



yO Me M 



° Me Me 

CO;, 



NH 



10 



p.p.m. 




CHO 



Like double bonds, cage structures pre- 
vent bond rotation, and can make the two 
protons of a CH2 group appear different. 
There are many flavouring compounds from 
herbs that have structures like this. In the 
example here — myrtenal, from the myrtle bush — there is a four- 
membered ring bridged across a six-membered ring. The CH 2 group 



myrtenal 



2.49 H 



1.33 Me 




Protons on saturated carbon atoms 



249 



on the bridge has two different hydrogen atoms — one is over a methyl group and the other is over 
the enal system. No rotation of any bonds in the cage is possible, so these hydrogens are always dif- 
ferent and resonate at different frequencies (1.04 and 2.49 p.p.m.). The methyl groups on the other 
bridge are also different for the same reason. 

Chemical shifts of CH groups 

A CH group in the middle of a carbon skeleton resonates at about 1.7 p.p.m. — another 0.4 p.p.m. 
downfield from a CH2 group. It can have up to three substituents and these will cause further down- 
field shifts of about the same amount as we have already seen for CH3 and CH2 groups. Here are three 
examples from nature: nicotine, the compound in tobacco that causes the craving (though not the 
death, which is doled out instead by the carbon monoxide and tars in the smoke), has one hydrogen 
atom trapped between a simple tertiary amine and an aromatic ring at 3.24 p.p.m. Lactic acid has a CH 
proton at 4.3 p.p.m.. You could estimate this with reasonable accuracy by taking 1.7 (for the CH) and 
adding 1.0 (for C=0) plus 2.0 (for OH) = 4.7 p.p.m. Vitamin C (ascorbic acid) has two CHs. One at 
4.05 p.p.m. is next to an OH group (estimate 1.7 + 2.0 for OH = 3.7 p.p.m.) and one next to a double 
bond and an oxygen atom at 4.52 p.p.m. (estimate 1.7 + 1 for double bond + 2 for OH = 4.7 p.p.m.). 



3.24 




1.41 




4.30 



OMe 



3.79 




2.17 



nicotine 



methyl ester 
of lactic acid 



4.52 

HO' OH 

vitamin C (ascorbic acid) 



An interesting case is the amino acid phenylalanine whose CH 2 group we looked at a moment 
ago. It also has a CH group between the amino and the carboxylic acid groups. If we record the H 
NMR spectrum in D 2 0, either in basic (NaOD) or acidic (DC1) solutions we see a large shift of that 
CH group. In basic solution the CH resonates at 3.60 p.p.m. and in acidic solution at 4.35 p.p.m. 
There is a double effect here: CO2H and NH3 are both more electron-withdrawing than CO2 and 
NH2 so both move the CH group downfield. 



C0 2 H pci 




D,0 



3.60 




4.35 



D 2 0, NaOD, and DCI have to be 
used in place of their 1 H 
equivalents to avoid swamping 
the spectrum with H 2 protons. 
All acidic protons are replaced by 
deuterium in the process — more 
on this later. 



phenylalanine 



Your simple guide to chemical shifts 

We suggest you start with a very 
simple (and therefore oversimpli- 
fied) picture, which should be the 
basis for any further refinements. 
Start methyl groups at 0.9, methyl- 
enes (CH2) at 1.3, and methines 
(CH) at 1.7 p.p.m. Any functional 
group is worth a one p.p.m. down- 
field shift except oxygen and halo- 
gen which are worth two p.p.m. 
This diagram summarizes the basic 
position. 



approximate guide Ct 
to group shifts in 
proton NMR spectra 1- 


H CH 

7 1. 


2 CH; 

3 O.i 


) 


-* -* — 








■« 

2 p.p.m. 




-< 

1 p.p.m. 






oxygen 

halogens 

nitro 

NCOR 




alkene, aryl 

carbonyl, nitrile 

sulfur 

nitrogen 















250 



11 ■ Proton nuclear magnetic resonance 



This is a very rough and ready guide and you can make it slightly more accurate by adding subdi- 
visions at 1.5 and 2.5 p.p.m. and including the very electron-withdrawing groups (nitro, ester, fluo- 
ride), which shift by 3 p.p.m. This gives us the summary chart on this page, which we suggest you use 
as a reference. 



If you want more detailed information, 
you can refer to the tables in Chapter 
15 or better still the more 
comprehensive tables in any 
specialized text. 



Summary chart of proton NMR shifts 

values to be added to 0.9 for CH 3 , 1.3 for CH 2 or 1.7 for CH 



shift 1 p.p.m. 



includes: 

aldehydes -CHO 

ketones -COR 

acids -C0 2 H 

esters -C0 2 R 

amides -C0NH 2 




N 





alkene 


C=C 


alkyne 


C=CR 


nitrile 


C=N 


carbonyl 


— c=o 


thiol 


SH 


sulfide 


SR 



includes: 

benzene -Ar 

heterocycles 

e.g. pyridine 



shift 1.5 p.p.m. 



^> 



aryl ring 

amine 

sulfoxide 



-Ar 
-NH 2 

-S— R 

II 




shift 2 p.p.m. 



includes: 
chloride -CI 
bromide -Br 

iodide -1 




> 





alcohol 

ether 

amide 

halide 

sulfone 



-OH 

-OR 

-NHCOR 

-Hal 

-S0 2 R 



shift 2.5 p.p.m. 



aryl ether 



-OAr 



shift 3 p.p.m. 




The alkene region and the benzene region 



251 



Answers deduced from this chart won't be very accurate but will give a good guide. Remember — 
these shifts are additive. Take a simple example, the ketoester below. There are just three signals and 
the integration alone distinguishes the two methyl groups from the CH2 group. One methyl has been 
shifted from 0.9 p.p.m. by about 1 p. p.m., the other by more than 2 p. p.m. The first must be next to 
C=0 and the second next to oxygen. More precisely, 2.14 p.p.m. is a shift of 1.24 p.p.m. from our 
standard value (0.9 p.p.m.) for a methyl group, about what we expect for a methyl ketone, while 3.61 
p.p.m. is a shift of 2.71 p.p.m., close to the expected 3.0 p.p.m. for an ester joined through the oxygen 
atom. The CH2 group is next to an ester and a ketone carbonyl group and so we expect it at 1.3 + 1.0 
+ 1.0 = 3.3 p.p.m., an accurate estimate, as it happens. We shall return to these estimates when we 
look at spectra of unknown compounds. 



CH 



o 

H H 



f 



-J 



10 



p.p.m. 



The alkene region and the benzene region 

In C NMR, one region was enough for both of these, but see how different things are 
with proton NMR. 

The two carbon signals are almost the same (1.3 p.p.m. difference < 1% of the total 
200 p.p.m. scale) but the proton signals are very different (1.6 p.p.m. difference = 16% of 
the 10 p.p.m. scale). There must be a fundamental reason for this. 

The benzene ring current causes large shifts for aromatic protons 

A simple alkene has an area of low electron density in the plane of the molecule because 
the 71 orbital has a node there, and the carbons and hydrogen nuclei lying in the plane gain no shield- 
ing from the Tt electrons. 

The benzene ring looks similar at first sight, and the plane of the molecule is indeed a node for all 
the Jt orbitals. However, benzene is 'aromatic' — it has extra stability because the six % electrons fit 
into three very stable orbitals and are delocalized round the whole ring. 

The applied field sets up a ring current in these delocalized electrons that produces a local field 
rather like the field produced by the electrons around a nucleus. Inside the benzene ring, the induced 
field opposes the applied field but, outside the ring, it reinforces the applied field. The carbon atoms 
are in the ring itself and experience neither effect, but the hydrogens are outside the ring, feel a 
stronger applied field, and appear less shielded. 





"Y^l 


C*\ 




JO 


u 


13 C shift, p.p.m. 


127.2 


128.5 


1 H shift, p.p.m. 


5.68 


7.27 




ring current 



applied 

magnetic 

field 



<§® 



induced field 




nodal plane 



Chapter 7 was devoted to a discussion 
of aromaticity and derealization. 



Magnetic fields produced by 
circulating electrons are all 
around you: electromagnets and 
solenoids are exactly this. 



252 



11 ■ Proton nuclear magnetic resonance 



Cyclophanes and annulenes 

You may think that it is rather pointless imagining 
what goes on inside an aromatic ring as we cannot 
have hydrogen atoms literally inside a benzene ring. 
However, we can get close. Compounds called 
cyclophanes have loops of saturated carbon atoms 
attached at both ends to the same benzene rings. 
You see here a structure for [7]para-cyclophane, 
which has a string of seven CH2 groups attached to 
the para positions of the same benzene ring. The four 
protons on the benzene ring itself appear as one line 
at a normal 8 7.07 p.p.m. The two CH2 groups joined 
to the benzene ring (CI) are deshielded by the ring 
current at 8 2.64 p.p.m. The next two sets of CH2 
groups on C2 and C3 are neither shielded nor 
deshielded at 8 1.0 p.p.m. The middle CH2 group in 
the chain (C4) must be pointing towards the ring in 
the middle of the n system and is heavily shielded by 
the ring current at negative 8 (-0.6 p.p.m.). 




8„1.0 



5 


S„1.0 


Sh-0.6 






H 'H 


3 


8 H 2.84 \ 
H 


£ ^— 



[7]-para-cyclophane 

With a larger aromatic ring, it is possible actually to 
have hydrogen atoms inside the ring. Compounds 
are aromatic if they have 4n + 2delocalized 
electrons and this ring with nine double bonds, that 
is, 18 it electrons, is an example. The hydrogens 
outside the ring resonate in the aromatic region at 
rather low field (9.28 p.p.m.) but the hydrogen 



atoms inside the ring resonate at an amazing -2.9 
p.p.m. showing the strong shielding by the ring 
current. Such extended aromatic rings are called 
annulenes: you met them in Chapter 7. 

Hs outside the ring Sh +9.28 p.p.m. 




Uneven electron distribution in aromatic rings 







J 




II 



10 



p.p.m. 



The NMR spectrum of this simple aromatic amine has three peaks in the ratio 1:2:2 which must be 
3H:6H:6H. The 6.38 p.p.m. signal clearly belongs to the protons round the benzene ring, but why are 
they at 6.38 and not at 7.27 p.p.m.? We must also distinguish the two methyl groups at 2.28 p.p.m. 
from those at 2.89 p.p.m. The chart on p. 000 suggests that these should both be at about 2.4 p.p.m., 
close enough to 2.28 p.p.m. but not to 2.89 p.p.m. The solution to both these puzzles is the distribu- 
tion of electrons in the aromatic ring. Nitrogen feeds electrons into the 71 system making it electron- 
rich: the ring protons are more shielded and the nitrogen atom becomes positively charged and its 
methyl groups more deshielded. The peak at 2.89 p.p.m. belongs to the NMe2 group. 





lone pair in 

p orbital 
on nitrogen 



Other groups, such as simple alkyl groups, hardly perturb the aromatic system at all and it is quite 
common for all five protons in an alkyl benzene to appear as one signal instead of the three we might 
expect. Here is an example with some nonaromatic protons too: there is another on p. 000 — the 
Cbz-protected amino acid. 



The alkene region and the benzene region 



253 



OMe 




10 



p. p.m. 



The five protons on the aromatic ring all have the same chemical shift. The OCH3 group is typical of 
a methyl ester (the chart on p. 000 gives 3.9 p. p.m.). One CH2 group is between two carbonyl groups 
(cf. 8 3.35 p.p.m. for the similar CH2 group on p. 000). The other is next to an ester and a benzene ring: 
we calculate 1.3 + 1.5 + 3.0 = 5.8 p.p.m. for that — reasonably close to the observed 5.19 p.p.m. 

How electron donation and withdrawal change chemical shifts 

We can get an idea of the effect of electron distribution by looking at a series of 1,4-disubstituted 
benzenes. This pattern makes all the remaining hydrogens in the ring the same. The compounds are 
listed in order of chemical shift: largest shift (lowest field) first. Benzene itself resonates at 7.27 p.p.m. 
Conjugation is shown by the usual curly arrows, and inductive effects by a straight arrow by the side 
of the group. Only one effect and one hydrogen atom are shown; in fact, both groups exert the same 
effect on all four identical hydrogen atoms. 
electron-withdrawing groups 



by conjugation 




O^® ^O © 



O^H 



by inductive effect 




The largest shifts come from groups that withdraw electrons by conjugation. Nitro is the most 
powerful — this should not surprise you as we saw the same in nonaromatic compounds both in C 
and H NMR spectra. Then come the carbonyl groups and nitrile followed by the few groups show- 
ing simple inductive withdrawal. CF3 is an important example of this kind of group — three fluorine 
atoms combine to exert a powerful effect. 

electron-donating and -withdrawing groups 



balance between withdrawal by inductive effect and donation of lone pairs by conjugation 




H 7.40 




H 7.32 




H 7.24 




H 7.00 



Conjugation, as discussed in 
Chapters 7 and 10, is felt through 
it bonds, while inductive effects 
are the effects of electron 
withdrawal or donation felt simply 
by polarization of the a bonds of 
the molecule. See p. 000. 



254 



11 ■ Proton nuclear magnetic resonance 



In the middle, around the position of benzene itself at 5 7.27 p. p.m., come the halogens whose 
inductive electron withdrawal and lone pair donation are nearly balanced. 



This all has very important 
consequences for the reactivity of 
differently substituted benzene 
rings: their reactions will be 
discussed in Chapter 22. 



Proton NMR is, in fact, a better 
guide to the electron density at 
carbon than is carbon NMR. 



In Chapter 10 we used ld C NMR 
to convince you that a carbonyl 
group polarized a conjugated 
alkene; we hope you find the -"-H 
NMR data even more convincing. 
Conjugate addition occurs to 
those very atoms whose electron 
deficiency we can measure by 
proton NMR. 



electron-donating groups 




balance between withdrawal by inductive effect and donation 
of lone pairs by conjugation; electron donation wins: 



7.03 




7.03 





Element 


Electronegativity 


5 H , p.p.m. 


Shift from 
7.27 


F 


4.1 


7.00 


-0.27 





3.5 


6.59 


-0.68 


N 


3.1 


6.35 


-0.92 



Alkyl groups are weak inductive donators and at the smallest shift we have the groups that, on bal- 
ance, donate electrons to the ring and increase the shielding at the carbon atoms. Amino is the best of 
these. So a nitrogen-based functional group (NO2) is the best electron withdrawer while another 
(NH2) is the best electron donor. 

As far as the donors with lone pairs are concerned, two factors are important — the size of the lone 
pairs and the electronegativity of the element. If we look at the four halides (central box above) the 
lone pairs are in 2p(F), 3p(Cl), 4p(Br), and 5p(I) orbitals. In all cases the orbitals on the benzene ring 
are 2p so the fluorine orbital is of the right size and the others too large. Even though fluorine is the 
most electronegative, it is still the best donor. 

Now comparing the groups in the first 
row of the p block elements. F, OH, NH2, 
all have lone pairs in 2p orbitals so electro- 
negativity is the only variable. As you 
would expect, the most electronegative 
element, F, is now the weakest donor. 

Electron- rich and electron-deficient alkenes 

The same sort of thing happens with alkenes. We'll concentrate on cyclohexene so as to make a good 
comparison with benzene. The six identical protons of benzene resonate at 7.27 p.p.m.; the two iden- 
tical alkene protons of cyclohexene resonate at 5.68 p.p.m. A conjugating and electron-withdrawing 

group such as a ketone removes electrons from the double bond as 
expected — but unequally. The proton nearer the C=0 group is only 
slightly downfield from cyclohexene but the more distant one is over 
1 p.p.m. downfield. The curly arrows show the electron distribution, 
which we can deduce from the NMR spectrum. 

Oxygen as a conjugating electron donor is even more dramatic. It 
shifts the proton next to it downfield by the inductive effect but 
pushes the more distant proton upfield again by a whole p.p.m. by 
donating electrons. The separation between the two protons is 
nearly two p.p.m. 

For both types of substituent, the effects are more marked on the 
more distant ((3) proton. If these shifts reflect the true electron distribution, we can deduce that 
nucleophiles will attack the electron-deficient site in the nitroalkene, while electrophiles will be 
attacked by the electron-rich sites in silyl enol ethers and enamines. These are all important reagents 
and do indeed react as we predict, as you will see in later chapters. Look at the difference — there are 
nearly 3 p.p.m. between the nitro compound and the enamine! 




7.27 


Cx 


5.68 


7.27 


KJ 


5.68 


6.0 


0\ 


4.65 


7.0 


"£ 


6.35 



The aldehyde region: unsaturated carbon bonded to oxygen 



255 



I© 




v o e 



H 7.31 



a 




-SiMe 3 



H 4.73 




Me 

I 
N. : 



~Me 



H 4.42 



electron-deficient 
nitroalkene 



electron-rich 
silyl enol ether 



electron-rich 
enamine 



Structural information from the alkene region 

Alkene protons on different carbon atoms can obviously be different if the carbon atoms themselves 
are different and we have just seen examples of that. Alkene protons can also be different if they are 
on the same carbon atom. All that is necessary is that the substituents at the other end of the double 
bond should themselves be different. The silyl enol ether and the unsaturated ester below both fit 
into this category. The protons on the double bond must be different, because each is cis to a differ- 
ent group. The third compound is an interesting case: the different shifts of the two protons on the 
ring prove that the N-Cl bond is at an angle to the C=N bond. If it were in line, the two hydrogens 
would be identical. The other side of the C=N bond is occupied by a lone pair and the nitrogen atom 



is trigonal (sp hybridized). 
silyl enol ether 

OSiMe 3 

Me' 



unsaturated ester 



chloroimine 




COoMe 



Me 

1.95 (3H) 




H 6.10 



7.50 H 



H 5.56 




7.99 



DMF is similar: as we saw earlier 
(p. 000), it has two different 
methyl groups because of the 
double bond. 



The aldehyde region: unsaturated carbon bonded to oxygen 

The aldehyde proton is unique. It is directly attached to a carbonyl group — one of the most electron- 
withdrawing groups that exists — and is very deshielded, resonating with the largest shifts of any CH 
protons in the 9-10 p.p.m. region. The examples below are all compounds that we have met before. 
Two are just simple aldehydes — aromatic and aliphatic. The third is the solvent DMF. Its CHO pro- 
ton is less deshielded than most — the amide derealization that feeds electrons into the carbonyl 
group provides some extra shielding. 




SMe 



10.14 



an aromatic aldehyde 




Me 

I 



Me' 



9.0 H 

an alphatic aldehyde 



8.01 H 
DMF 



Conjugation with an oxygen atom has much the same effect — formate esters resonate at about 8 
p.p.m. — but conjugation with 7t bonds does not. The simple conjugated aldehyde below and myrte- 
nal both have CHO protons in the normal region (9-10 p.p.m.). 



-8.0 



n 
R -nA r 



a formate ester 



H 9.95 




H 5.88 
3-methylbut-2-enal 




myrtenal 



Aliphatic is a catch-all term for 
compounds that are not aromatic. 



256 



11 ■ Proton nuclear magnetic resonance 



Two other types of protons resonate in this region: some aromatic protons and some protons 
attached to heteroatoms like OH and NH. The first of these will provide our discussion on structural 
information and the second will be the subject of the section following that discussion. 



O. 



a 



"0 © 



> 



H 7.31 

electron-deficient 
nitroalkene 

0® 

I® 

H 8.48 
1,4-dinitrobenzene 




Please note that the alternative 
'conjugation' shown in this figure 
is wrong. The structure with two 
adjacent double bonds in a six- 
membered ring is impossible and, 
in any case, as you saw in 
Chapter 8, the lone pair electrons 
on nitrogen are in an sp 2 orbital 
orthogonal to the p orbitals in the 
ring. There is no interaction 
between orthogonal orbitals. 

X N H N H 

M) ® 



incorrect 
derealization 



impossible 
structure 



Structural information from the aldehyde region 

Protons on double bonds, even very electron-deficient double bonds like those of nitroalkenes, 
hardly get into the aldehyde region. However, some benzene rings with very electron-withdrawing 
groups do manage it because of the extra downfield shift of the ring current, so beware of nitroben- 
zenes as they may have signals in the 8-9 p. p.m. region. 

More important molecules with signals in this region are the aromatic heterocycles such as 
pyridine, which you met in Chapter 7. The NMR shifts clearly show that pyridine is aromatic and we 
discussed its basicity in Chapter 8. One proton is at 7.1 p.p.m., essentially the same as benzene, but 
the others are more downfield and one, at C2, is in the aldehyde region. This is not because pyridine 
is 'more aromatic' than benzene but because nitrogen is more electronegative than carbon. Position 
C2 is like an aldehyde — a proton attached to sp C bearing a heteroatom — while C4 is electron- 
deficient by conjugation (the electronegative nitrogen is electron-withdrawing). Isoquinoline is a 
pyridine and a benzene ring fused together and has a proton even further downfield at 9.1 p.p.m. — 
this is an imine proton that experiences the ring current of the benzene ring. 
H H 7.5 H 



H 7.1 






conjugation in pyridine 




H 8.5 



isoquinoline 



Protons on heteroatoms are more variable than protons on carbon 

Protons directly attached to O, N, or S (or any other heteroatom, but these are the most important) 
also have signals in the NMR spectrum. We have avoided them so far because the positions of these 
signals are less reliable and because they are affected by exchange. 




r 

L 



10 



p.p.m. 



In Chapter 3 we looked at the C NMR spectrum of BHT. Its proton NMR is very simple, con- 
sisting of just four lines with integrals 2, 1, 3, and 18. The chemical shifts of the f-butyl group, the 
methyl group on the benzene ring, and the two identical aromatic protons should cause you no sur- 
prise. What is left, the 1H signal at 5.0 p.p.m., must be the OH. Earlier on in this chapter we saw the 
spectrum of acetic acid CH3CO2H, which showed an OH resonance at 11.2 p.p.m. Simple alcohols 
such as f-butanol have OH signals in CDCI3 (the usual NMR solvent) at around 2 p.p.m. Why such 
differences? 



The aldehyde region: unsaturated carbon bonded to oxygen 



257 



u 

t^^, 



2 
CH; 



11.2 
H 



y 



1.28 
CH 3 
CH 3 



CH, 



1.91 



1.42 
CH 3 
CH 3 



CH 3 

A. 



1.82 



1.15 
CH; 

CH 



CH 3 

A. 



H 

1.20 



H 



acetic acid 



t-BuSH in CDCU 



t-BuNH 2 in CDCI 3 



Functional group 

5 H (0H), p.p.m. 



f-BuOH in CDCI3 

This is a matter of acidity. The more acidic a proton is — that is, the more 
easily it releases H (this is the definition of acidity from Chapter 8) — the 
more the OH bond is polarized towards oxygen. The more the RO-H bond is 
polarized, the closer we are to free H + , which would have no shielding elec- 
trons at all, and so the further the proton goes downfield. The OH chemical 
shifts and the acidity of the OH group are very roughly related. 

Thiols (RSH) behave in a similar way to alcohols but are not so deshielded, as you would expect 
from the smaller electronegativity of sulfur (phenols are all about 5.0 p.p.m., PhSH is at 3.41 p.p.m.). 
Alkane thiols appear at about 2 p.p.m. and arylthiols at about 4 p.p.m. Amines and amides show a 
big variation, as you would expect for the variety of functional groups involved, and are summarized 
below. Amides are slightly acidic, as you saw in Chapter 8, and amide protons resonate at quite low 
fields. Pyrroles are special — the aromaticity of the ring makes the NH proton unusually acidic and 
they appear at about 10 p.p.m. 
chemical shifts of NH protons 



Alcohol 
ROH 

16 
2.0 



Alkyl — NH 2 Aryl 



§NH ~ 3 



— nh, Jk.„s* „J\.,/ Alk y „A^,/ Ar y \„/ 



*N 

I 
H 

§NH ~5 



N 
H 



-10 



-10 



Exchange of acidic protons is revealed in proton NMR spectra 

Compounds with very polar groups often dissolve best in water. NMR spectra are usually run in 
CDCI3, but heavy water, D 2 0, is an excellent NMR solvent. Here are some results in that medium. 



KUN 



CO;H 



H3N 



CO, 



amino 31 ■ ' zwifterkin 


( 




the simplest amino acid glycine 


I 






^=k_ 


— \ 




r 




the salt of an amino thiol 


A 





Phenol 
ArOH 

10 
5.0 



Carboxylic acid 
RC0 2 H 

5 
>10 



2NH,, 



^ ISM 

diamnwnium EDTA 




EDTA is ethylenediamine 
tetraacetic acid, an important 
complexing agent for metals. This 
is the salt formed with just two 
equivalents of ammonia. 



p.p.m. 



258 11 ■ Proton nuclear magnetic resonance 



Glycine is expected to exist as a zwitterion (Chapter 8, p. 000). It has a 2H signal for the CH 2 
between the two functional groups, which would do for either form. The 3H signal at 4.90 p. p.m. 
might suggest the NH3 group, but wait a moment before making up your mind. The aminothiol salt 
has the CMe2 and CH2 groups about where we would expect them, but the SH and NH3 protons 
appear as one 4H signal. The double salt of EDTA has several curious features. The two CH2 groups 
in the middle are fine, but the other four CH2 groups all appear identical as do all the protons on 
both the C0 2 H and NH3 groups. 

The best clue to why this is so involves the chemical shifts of the OH, NH, and SH protons in 
these molecules. They are all the same within experimental error: 4.90 p. p.m. for glycine, 4.80 p. p.m. 
for the aminothiol, and 4.84 p.p.m. for EDTA. They all correspond to the same species: HOD. 
Exchange between XH (where X = O, N, or S) protons is extremely fast, and the solvent, D 2 0, sup- 
plies a vast excess of exchangeable deuteriums. These immediately replace all the OH, NH, and SH 
protons in the molecules with D, forming HOD in the process. Recall that we do not see signals 
for deuterium atoms (that's why deuterated solvents are used). They have their own spectra at a 
different frequency. 

® ^^ 

+ DOH 




D 2 »► yr ^N + OH 




'N N' ^C0 2 G ®0 2 C^^N N'"'~ X "C0 2 G 

) k " ) k 

H0 2 (T N C0 2 H D0 2 (T N C0 2 H 

2 NH 4 ® + D 2 2 NH 4 ® + DOH 

The same sort of exchange between OH or NH protons with each other or with traces of water in 
the sample means that the OH and NH peaks in most spectra in CDCI3 are rather broader than the 
peaks for CH protons. 

Two questions remain. First, can we tell whether glycine is a zwitterion in water or not? Not really: 
the spectra fit either or an equilibrium between both. Other evidence leads us to prefer the zwitterion 
in water. Second, why are all four CH2CO groups in EDTA the same? This we can answer. As well as 
the equilibrium exchanging the CO2H protons with the solvent, there will be an equally fast equilibri- 
um exchanging protons between CO2H and CO2D. This makes all four 'arms' of EDTA the same. 

You should leave this section with an important chemical principle firmly established in your 
mind. 

1 Protons exchange fast 

Proton exchange between heteroatoms, particularly O, N, and S, is a very fast 
process in comparison with other chemical reactions, and often leads to averaged 
peaks in the H NMR spectrum. 



You will need this insight as you study organic mechanisms. 

Coupling in the proton NMR spectrum 

Nearby hydrogen nuclei interact and give multiple peaks 

So far proton NMR has been not unlike carbon NMR on a smaller scale. However, we have yet to 
discuss the real strength of proton NMR, something more important than chemical shifts and 



Coupling in the proton NMR spectrum 



259 



something that allows us to look not just at individual atoms but also at the way the C-H skeleton 
is joined together. This is the result of the interaction between nearby protons known as 
coupling. 

An example we could have chosen in the last section is the nucleic acid component, cytosine, 
which has exchanging NH2 and NH protons giving a peak for HDO at 4.5 p. p.m. We didn't choose 
this example because the other two peaks would have puzzled you. Instead of giving just one line 
each, they give two lines each — doublets as you will learn to call them — and it is time to discuss the 
origin of this 'coupling'. 



60 MHz J H NMR spectrum 
of cytosine in D 2 






HOD 



Cytosine is one of the four bases 
that, in combination with 
deoxyribose and phosphate, 
make up DNA. It is a member of 
the class of heterocycles called 
pyrimidines. We come back to the 
chemistry of DNA towards the end 
of this book, in Chapter 49. 



1 
- 



8 



3 p. p.m. 



You might have expected a spectrum like that of the heterocycle below, which is also a pyrimi- 
dine. It too has exchanging NH2 protons and two protons on the heterocyclic ring. But these two 
protons give the expected two lines instead of the four lines in the cytosine spectrum. It is easy to 
assign the spectrum: proton H is attached to an aldehyde-like C=N and so comes at lowest field. 
The proton H is ortho to two electron-donating NH2 groups and so comes at high field for an aro- 
matic proton (p. 000). These protons do not couple with each other because they are too far apart. 
They are separated by five bonds whereas the ring protons in cytosine are separated by just three 
bonds. 

60 MHz J H NMR spectrum 

of 4,6-diaminopyrimidine in D 2 



J 



I 





rr "nh. 



HOD 



6 



3 p. p.m. 



Understanding this phenomenon is so important that we are going to explain it in three different 
ways — you choose which appeals to you most. Each method offers a different insight. 

The pyrimidine spectrum has two single lines (singlets we shall call them from now on) because 
each proton, H or H , can be aligned either with or against the applied magnetic field. The cytosine 
spectrum is different because each proton, say, H , is near enough to experience the small magnetic 
field of the other proton H as well as the field of the magnet itself. The diagram shows the 
result. 



260 



11 ■ Proton nuclear magnetic resonance 



spectrum with 
no interaction 



effect of H x and 

applied field acting 

together on H A 



t 



H A 



A 

7.5 



A 

7.5 



H x 



A H 
I ai 



x aligned with 
pplied field 



effect of H x and 
applied field acting 
in opposition on H A 



t 



. 



H aligned against 
" applied field 



A 

7.5 



effect of H x on H A and 
H A on H x 



T 



T 



resultant 
spectrum 



7-5 5.8 

If each proton interacted only with the applied field we would get two singlets. But proton H 
actually experiences two slightly different fields: the applied field plus the field of H or the applied 
field minus the field of H . H acts either to increase or to decrease the field experienced by H . The 
position of a resonance depends on the field experienced by the proton so these two situations give 
rise to two slightly different peak — a doublet as we shall call it. And whatever happens to H happens 
to H as well, so the spectrum has two doublets, one for each proton. Each couples with the other. The 
field of a proton is a very small indeed in comparison with the field of the magnet and the separation 
between the lines of a doublet is very small. We shall discuss the size of the coupling later (p. 000). 

The second explanation takes into account the energy levels of the nucleus. In Chapter 4, when we 
discussed chemical bonds, we imagined electronic energy levels on neighbouring atoms interacting 
with each other and splitting to produce new molecular energy levels, some higher in energy and 
some lower in energy than the original atomic energy levels. When hydrogen nuclei are near each 
other in a molecule, the nuclear energy levels also interact and split and produce new energy levels. If 
a single hydrogen nucleus interacts with a magnetic field, we have the picture on p. 000 of this chap- 
ter: there are two energy levels as the nucleus can be aligned with or against the applied magnetic 
field, there is one energy jump possible, and there is a resonance at one frequency. This you have now 
seen many times and it can be summarized as shown below. 



energy levels for one isolated nucleus H A 



A 



applied 

magnetic 

field 



nucleus A aligned 
magnetic field 



© 



nucleus A aligned 
magnetic field 




higher 
energy 
level 



lower 

energy 

level 



energy 



Coupling in the proton NMR spectrum 



261 



The spectrum of the pyrimidine on p. 000 showed two protons each independently in this 
situation. Each had two energy levels, each gave a singlet, and there were two lines in the 
spectrum. But, in the cytosine molecule, each proton has another hydrogen nucleus nearby 
and there are now four energy levels. Each nucleus H and H can be aligned with or against 
the applied field. There is one most stable energy level where they are both aligned with the field 
and one least stable level where they are both aligned against. In between there are two different 
energy levels in which one nucleus is aligned with the field and one against. Exciting H from 
alignment with to alignment against the applied field can be done in two slightly different ways, 
shown as Aj and A2 on the diagram. The result is two resonances very close together in the 
spectrum. 



energy levels for two interacting nuclei H A and H x 



both nuclei 
A and X aligned 
against applied 
magnetic field 



A 



X 2 : energy required 
to excite X with A 
but against B 



dXD 



A 2 : energy required 

to excite A with X 

but against 8 



nucleus A aligned 
against applied 
magnetic field: 
X aligned with 



©©■ 
V 



these have slightly 
different energies 



0© 



nucleus A aligned 


with applied 


magnetic field: 


X aligned against 



applied 

magnetic 

field B 



■' 

Ai: energy required 


to excite A against X 


and against 6 



0© 



Xi: energy required 

to excite X against A 

and against 8 



both nuclei 
A and X aligned 
with applied 
magnetic field 



Please notice carefully that we cannot have this discussion about H without discussing H in the 
same way. If there are two slightly different energy jumps to excite H , there must also be two slight- 
ly different energy jumps to excite H . The difference between Ai and A2 is exactly the same as the 
difference between Xi and X2. Each proton now gives two lines (a doublet) in the NMR spectrum 
and the splitting of the two doublets is exactly the same. We describe this situation as coupling. We 
say 'A and X are coupled' or 'X is coupled to A' (and vice versa, of course). We shall be using this lan- 
guage from now on and so must you. 

Now look back at the spectrum of cytosine at the beginning of this section. You can see the 
two doublets, one for each of the protons on the aromatic ring. Each is split by the same amount 
(this is easy to check with a ruler) and the separation of the lines is the coupling constant and 
is called /. In this case / = 4 Hz. Why do we measure / in hertz and not in p. p.m.? We measure 
chemical shifts in p. p.m. because we get the same number regardless of the rating of the NMR 
machine in MHz. We measure / in Hz because we also get the same number regardless of the 
machine. 



262 



11 ■ Proton nuclear magnetic resonance 



► Measuring coupling 
constants in hertz 

To measure a coupling constant it 
is essential to know the rating of 
the NMR machine in MHz 
(MegaHertz). This is why you are 
told that each illustrated 
spectrum is, say, a '250 MHz 1 H 
NMR spectrum'. To measure the 
coupling, measure the distance 
between the lines by ruler or 
dividers and use the horizontal 
scale to find out the separation in 
p. p.m. The conversion is then 
easy — to turn parts per million of 
megahertz into hertz you just 
leave out the million! So 1 p. p.m. 
on a 300 MHz machine is 300 Hz. 
On a 90 MHz machine it would be 
90 Hz. 




!? :ll. 



-j - a,8 Hi 



spectrum recorded on 
12.3 Hi— - 90 MHj spectrometer 




8 






J 



- - 12.3 rt! 



• 



spectrum recorded on 



12.3 Hz J . 300 MHi spectrometer 






.lp.p-m. = SOP Hi 



3 p.p.m. 



• Spectra from different machines 

When you change from one machine to another, say, from an 80 MHz to a 500 
MHz NMR machine, chemical shifts (8) stay the same in p.p.m. but coupling 
constants (/) stay the same in Hz. 

Now for the third way to describe coupling. If you look again at what the spectrum would be like 
without interaction between H and H you would see this, with the chemical shift of each proton 
clearly obvious. 



spectrum of molecule without coupling 



7.5 



5.8 



But you don't see this because each proton couples with the other and splits its signal by an equal 
amount either side of the true chemical shift. 



the two protons couple: 
H A 

X 


H x 

X 


AA 


J AX ** *" 4 Hz 4 Hz -» »- 



The true spectrum has a pair of doublets each split by an identical amount. Note that no line 
appears at the true chemical shift, but it is easy to measure the chemical shift by taking the midpoint 
of the doublet. 



Coupling in the proton NMR spectrum 



263 



spectrum with coupling 

5 A = 7.5 p. p.m. 

I t | 



8x = 5.8 p. p.m. 



Ax = 4 Hz 



Jxa = 4 Hz 



So this spectrum would be described as 8 H 7.5 (1H, d, /4 Hz, H A ) and 5.8 (1H, d, ]4 Hz, H x ). 
The main number gives the chemical shift in p.p.m. and then, in brackets, comes the integra- 
tion as the number of Hs, the shape of the signal (here 'd' for doublet), the size of coupling con- 
stants in Hz, and the assignment, usually related to a diagram. The integration refers to the 
combined integral of both peaks in the doublet. If the doublet is exactly symmetrical, each peak inte- 
grates to half a proton. The combined signal, however complicated, integrates to the right number of 
protons. 

We have described these protons as A and X with a purpose in mind. A spectrum of two equal 
doublets is called an AX spectrum. A is always the proton you are discussing and X is a proton with a 
very different chemical shift. The alphabet is used as a ruler: nearby protons (on the chemical shift 
scale — not necessarily nearby in the structure!) are called B, C, etc. and distant ones are called X, Y, 
etc. You will see the reason for this soon. 

If there are more protons involved, the splitting process continues. Here is the NMR spectrum of 
a famous perfumery compound supposed to have the smell of 'green leaf lilac'. The compound is an 
acetal with five nearly identical aromatic protons at the normal benzene position (7.2-7.3 p.p.m.) 
and six protons on two identical OMe groups. 



r 




OMe 



OMc 



90 MHz 



Aa_ 



=1 



H ft 




r 



JiAJL 



3 



L 



It is the remaining three protons that interest us. They appear as a 2H doublet at 2.9 p.p.m. 
and a 1H triplet at 4.6 p.p.m. In NMR talk, triplet means three equally spaced lines in the ratio 
1:2:1. The triplet arises from the three possible states of the two identical protons in the CH2 
group. 

If one proton H interacts with two protons H , it can experience three states of proton 
H . Both protons H can be aligned with the magnet or both against. These states will in- 
crease or decrease the applied field just as before. But if one proton H is aligned with, and 
one against the applied field, there is no net change to the field experienced by H and there 
are two possibilities for this (see diagram). We therefore see a signal of double intensity for H at 
the correct chemical shift, one signal at higher field and one at lower field. In other words, a 1:2:1 
triplet. 



264 



11 ■ Proton nuclear magnetic resonance 



2H* 



spectrum with 
no interaction 



effect of H x and 

applied field acting 

together on H A 



two H x s cancel out 

double intensity 

signal at true position 



effect of H x and 
applied field acting 
in opposition on H A 






4.6 
H A 



2H* 



both H x s aligned 
with applied field 



A 

4.6 

H A 



2H* 



|| and tw 



one H x aligned with applied field 
and one H x against (two ways) 



A 

4.6 
H A 



2H X 



both H x s aligned 
against applied field 



A 

4.6 
H A 



2H X 



effect of H x on H A 

and of H A on x 



resultant 
spectrum 



4.6 



2.9 



We could look at this result by our other methods too. There is one way in which both nuclei can 
be aligned with and one way in which both can be aligned against the applied field, but two ways in 
which they can be aligned one with and one against. Proton H interacts with each of these states. 
The result is a 1:2:1 triplet. 



applied 
field B„ 



two states 
of H A 



© 
© 



three states 
of 2 x H x 

\ ± X 2 

GXD 
GXD©© 

"1 "2 

©© 



Coupling in the proton NMR spectrum 



265 



Using our third way to see how the triplet arises, we can look at the splitting as it happens. 



coupling in an AX 2 system „a 



2xH x 



coupling to first H 



coupling to second H x 




Ax 
5 Hz 



coupling to H 



5 Hz 



JxA 

5 Hz 



the resulting AX 2 spectrum 




4 


4 




2 










1 




1 





















If there are more protons involved, we continue to get more complex systems, but the intensities 
can all be deduced simply from Pascal's triangle, which gives the coefficients in a binomial expan- 
sion. If you are unfamiliar with this simple device, here it is. 

Pascal's triangle 




^ Constructing Pascal's 
triangle 

Put '1' at the top and then add an 
extra number in each line by 
adding together the numbers on 
either side of the new number in 
the line above. If there is no 
number on one side, that counts 
as a zero, so the lines always 
begin and end with '1'. 



You can read off from the triangle what pattern you may expect when a proton is coupled to n 
equivalent neighbours. There are always n + 1 peaks with the intensities shown by the triangle. So far, 
you've seen 1:1 doublets (line 2 of the triangle) from coupling to 1 proton, and 1:2:1 triplets (line 3) 
from coupling to 2. You will often meet ethyl groups (CH3-CH2X) where the CH2 group appears as 
a 1:3:3:1 quartet and the methyl group as a 1:2:1 triplet and isopropyl groups (Clr^^CHX where the 
methyl groups appear as a 6H doublet and the CH group as a septuplet. The outside lines of a septu- 
plet are so weak (l/20th of the middle line) that it is often mistaken for a quintet. Inspection of the 
integral should put you on the right track. 

Here is a simple example, the four-membered cyclic ether oxetane. Its NMR spectrum has a 4H 
triplet for the two identical CH2 groups next to oxygen and a 2H quintet for the CH2 in the middle. 
Each proton H 'sees' four identical neighbours (H ) and is split equally by them all to give a 



266 



11 ■ Proton nuclear magnetic resonance 



► Constructing Pascal's 
triangle 

Remember, the coupling comes 
from the neighbouring protons: it 
doesn't matter how many protons 
form the signal itself (2 for H x , 4 
for H A ) — it's how many are next 
door (4 next to H x , 2 next to H A ) 
that matters. It's what you see 
that counts not what you are. 



1:4:6:4:1 quintet. Each proton H 'sees' two identical neighbours H and is split into a 1:2:1 triplet. 
The combined integral of all the lines in the quintet together is 2 and of all the lines in the triplet 
is 4. 



90 MHz 



■ 




HA H A 

oxetane 



-r- 
b 



3 



A slightly more complicated example is the diethyl acetal below. It has a simple AX pair 
of doublets for the two protons on the 'backbone' (red and green) and a typical ethyl group 
(2H quartet and 3H triplet). An ethyl group is attached to only one substituent through its 
CH2 group, so the chemical shift of that CH2 group tells us what it is joined to. Here the peak at 
3.76 p.p.m. can only be an OEt group. There are, of course, two identical CH2 groups in this 
molecule. 



y 



f 



j 



"Y 

a 



H 

90 Mi ■! 



So far, we have seen situations where a proton has several neighbours, but the coupling constants 
to all the neighbours have been the same. What happens when coupling constants differ? 
Chrysanthemic acid, the structural heart of the natural pyrethrin insecticides, gives an example of 
the simplest situation — where a proton has two different neighbours. 




chrysanthemic acid 
90 MHz 
^cOjtH x 3 expansion 

_A_A_ JJkX_ 



H* 



H* 



r 




r 



jujlZJI lb. 



Coupling in the proton NMR spectrum 



267 



This is an interesting three-membered ring compound produced by pyrethrum flowers (Chapter 
1). It has a carboxylic acid, an alkene, and two methyl groups on the three-membered ring. Proton 



jM 



H A has two neighbours, H A and H . The coupling constant to H A is 8 Hz, and that to H M is 5.5 Hz. 

The splitting pattern looks like this. 

The result is four lines of equal intensity 
called a double doublet (or sometimes a 
doublet of doublets), abbreviation dd. The 
smaller coupling constant can be read off 
from the separation between lines 1 and 2 
or between lines 3 and 4, while the larger 
coupling constant is between lines 1 and 3 
or between lines 2 and 4. You could see this 
as an imperfect triplet where the second 
coupling is too small to bring the central 
lines together: alternatively, look at a 
triplet as a special case of a double doublet 
where the two couplings are identical. 

Coupling is a through bond effect 

Neighbouring nuclei might interact through space or through the electrons in the bonds. We know 
that coupling is in fact a 'through bond effect' because of the way coupling constants vary with the 
shape of the molecule. The most important case occurs when the protons are at either end of a 
double bond. If the two hydrogens are cis, the coupling constant /is typically about 10 Hz but, if they 
are trans, /is much larger, usually 15-18 Hz. These two chloro acids are good examples. 

IT) H CI CI 

,C0 2 H ^^. .C0 2 H 



Abbreviations used for style of signal 


Abbreviation Meaning 


Comments 


s singlet 


might be 'broad' 


d doublet 


equal in height 


t triplet 


should be 1:2:1 


q quartet 


should be 1:3:3:1 


dt double triplet 


other combinations too, such 




as dd, dq, tq... 


m multiplet 


avoid if possible but 




sometimes necessary to 




describe complicated signals 



CI 




CI 



(J!) 





C0 2 H 



("J 




C0 2 H 



Ax 8 Hz 




8 Hz 



5.5 Hz 



Coupling constants to two 
identical protons must be 
identical but, if the protons differ, 
the coupling constants must also 
be different (though sometimes 
by only a very small amount). 



H atoms distant 
orbitals parallel 



hydrogens are trans 
J = 15 Hz 



H atoms close 
orbitals not parallel 



hydrogens are cis 
J=9 Hz 



If coupling were through space, the nearer cis hydrogens would have the larger /. In fact, coupling 
occurs through the bonds and the more perfect parallel alignment of the orbitals in the trans com- 
pound provides better communication and a larger /. 

Coupling is at least as helpful as chemical shift in assigning spectra. When we said that the protons 
on cyclohexenone had the chemical shifts shown, how did we know? It was coupling that told us the 
answer. The proton next to the carbonyl group has one neighbour and appears as a doublet with / = 
11 Hz, just right for a proton on a double bond with a cis neighbour. The proton at the other end 
appears as a double triplet. Inside each triplet the separation of the lines is 4 Hz and the two triplets 
are 11 Hz apart. This means the following diagramatically. 



peak heights 

shown on this 

triplet 



one line at 7.0 
with no coupling 



resultant double triplet for H 



11 Hz coupling to H 2 

4 Hz coupling 

~ to first H 4 



4 Hz coupling 
to second H 4 







11 Hz 










'T 1 




4 Hz 



For the same reason — orbital 
overlap — this anti arrangement of 
substituents is also preferred in 
chemical reactions such as 
elimination (Chapter 19) and 
fragmentation (Chapter 38). 
o 

l 2 5 6.0 



5 7.0 




5 6.0 



8 7.0 



268 



11 ■ Proton nuclear magnetic resonance 



This is what happens when a proton couples to different groups of protons with different coupling 
constants. Many different coupling patterns are possible, many can be interpreted, but others can- 
not. However, machines with high field magnets make the interpretation easier. As a demonstration, 
let us turn back to the bee alarm pheromone that we met in Chapter 3. An old 90 MHz NMR spec- 
trum of this compound looks like this. 



9 H H H H 



90 MHz 




You can see the singlet for the isolated black methyl group and just about make out the triplets for 
the green CH2 group next to the ketone (C3) at about 2.5 p.p.m. and for the orange methyl group at 
0.9 p.p.m. (C7) though this is rather broad. The rest is frankly a mess. Now see what happens when 
the spectrum is run on a more modern 500 MHz spectrometer. 




— r~ 
2,6 





2.0 



"~l 

p.p.m. 



2.4 



2.2 



1.8 



1.6 



1.4 



1.2 



1.0 



0,8 



Notice first of all that the chemical shifts have not changed. However, all the peaks have closed up. 
This is because / stays the same in Hz and the 7 Hz coupling for the methyl group triplet was 7/90 = 
0.07 p.p.m. at 90 MHz but is 7/500 = 0.014 p.p.m. at 500 MHz. In the high field spectrum you can 
easily see the singlet and the two triplets but you can also see a clear quintet for the red CH2 group at 
C4, which couples to both neighbouring CH2 groups with the same / (7 Hz). Only the two CH2 
groups at C5 and C6 are not resolved. However, this does not matter as we know they are therefrom 
the coupling pattern. The triplet for the orange methyl group at C7 shows that it is next to a CH2 
group, and the quintet for the CH2 group at C4 shows that it is next to two CH2 groups. We know 
about one of them, at C5, so we can infer the other at C6. 



Coupling in the proton NMR spectrum 



269 



Coupling constants depend on three factors 

In heptanone all the coupling constants were about the same but in cyclohexenone they were quite 
different. What determines the size of the coupling constant? There are three factors. 

• Through bond distance between the protons 

• Angle between the two C-H bonds 

• Electronegative substituents 

The coupling constants we have seen so far are all between hydrogen atoms on neighbouring car- 
bon atoms. The coupling is through three bonds (H-C-C-H) and is designated Thh- These cou- 
pling constants Thh are usually about 7 Hz in an open-chain, freely rotating system such as we have 
in heptanone. The C-H bonds vary little in length but the C-C bond might be a single or a double 
bond. In cyclohexenone it is a double bond, significantly shorter than a single bond. Couplings 
( 7hh) across double bonds are usually larger than 7 Hz (11 Hz in cyclohexenone). Thh couplings 
are called vicinal couplings because the protons concerned are on neighbouring carbon atoms. 

Something else is different too: in an open- chain system we have a time average of all rotational 
conformations. Across a double bond there is no rotation and the angle between the two C-H bonds 
is fixed because they are in the same plane. In the plane of the alkene, the C-H bonds are either at 60° 
(cis) or at 180° (trans) to each other. Coupling constants in benzene rings are slightly less than those 
across cis alkenes because the bond is longer (bond order 1.5 rather than 2). 



3 J HH coupling constants 

open chain 
single bond 



benzene ring 
longer bond (0.5 it bond) 



cis alkene 
double bond 



trans alkene 
double bond 







H 



free rotation 
J~ 7 Hz 



60° angle 
J 8-10 Hz 



60° angle 
J 10-12 Hz 



180° angle 
J 14-18 Hz 



In naphthalenes, there are unequal bond lengths around the two rings. The bond between the two 
rings is the shortest, and the lengths of the others are shown. Coupling across the shorter bond (8 
Hz) is significantly stronger than coupling across the longer bond (6.5 Hz). 

The effect of the third factor, electronegativity, is easily seen in the comparison between ordinary 
alkenes and enol ethers. We are going to compare two series of compounds with a cis or a trans 
double bond. One series has a phenyl group at one end of the alkene and the other has an OPh group. 
Within each box, that is for either series, the trans coupling is larger than the cis, as you would now 
expect. But if you compare the two series, the enol ethers have much smaller coupling constants. The 
trans coupling for the enol ethers is only just larger than the cis coupling for the alkenes. The 
electronegative oxygen atom is withdrawing electrons from the C-H bond in the enol ethers and 
weakening communication through the bonds. 

effect of electronegative substituents on 3 J HH - alkenes and enol ethers 





alkenes 


3 J cis = 11.5 Hz 3 J trans = 16.0 Hz 


H H 


H 

1 


)=( 


rV 


R Ph 


H 



enol ethers 


3 J cis = 6.0 Hz 


3 J trans = 12.0 Hz 


H H 


H 

1 


y/ 


A. .OPh 


/ \ 


R-^^^ 


R OPh 


H 



Conjugation in naphthalene was 
discussed in Chapter 7, p. 000. 




J 6.5 Hz 



naphthalene 



270 



11 ■ Proton nuclear magnetic resonance 



. . - J 9 Hz 

bond H 

order 




meta coupling 



rr 



< 4 J H h < 3 Hz 



allylic coupling 

XY 



Another good example is the coupling found in pyridines. Though the bond order is actually 
slightly less between C3 and C4, the coupling constants are about normal for an aromatic ring (com- 
pare naphthalene above), while coupling constants across C2 and C3, nearer to the electronegative 
nitrogen, are smaller. 

When the through bond distance gets longer, coupling is not usually seen. To put it another way, 
four-bond coupling Thh is usually zero. However, it is seen in some special cases, the most impor- 
tant being meta coupling in aromatic rings and allylic coupling in alkenes. In both, the orbitals 
between the two hydrogen atoms can line up in a zig-zag fashion to maximize interaction. This 
arrangement looks rather like a letter 'W and this sort of coupling is called W-coupling. Even with 
this advantage, values of Thh are usually small, about 1-3 Hz. 

Meta coupling is very common when there is ortho coupling as well, but here is an example where 
there is no ortho coupling because none of the aromatic protons have immediate neighbours — the 
only coupling is meta coupling. There are two identical H s, which have one meta neighbour and 
appear as a 2H doublet. Proton H between the two MeO groups has two identical meta neighbours 
and so appears as a 1H triplet. The coupling is small (/- 2.5 Hz). 





,OMe 



\r 



5 



4 p. p.m. 




H 2 5 6.0 



H 3 5 7.0 




no coupling 
between 
these identical 
neighbours 

one 4H singlet 



We have already seen a molecule with allylic coupling. We discussed in some detail why cyclo- 
hexenone has a double triplet for H . But it also has a less obvious double triplet for H . The triplet 
coupling is less obvious because / is small (about 2 Hz) because it is Thh — allylic coupling to the 
CH2 group at C4. Here is a diagram of the coupling, which you should compare with the earlier one 
for cyclohexenone. 



peak heights 

shown on this 

triplet 



one line at 6.0 
with no coupling 



1 1 



11 Hz coupling to H d 



/\ 



AA 



2 Hz allylic 
coupling 
to each H 4 



resultant double triplet for H 





11 Hz 






- 




4— 2 Hz 


large separation between triplets 



Coupling between similar protons 

We have already seen that identical protons do not couple with each other. The three protons in a 
methyl group may couple to some other protons, but never couple with each other. They are an A3 
system. Identical neighbours do not couple either. In the para-disubstituted benzenes we saw on p. 
000, all the protons on the aromatic rings were singlets. 



Coupling in the proton NMR spectrum 



271 



We have also seen how two different protons forming an AX system give two separate doublets. 
Now we need to see what happens to protons in between these two extremes. What happens to two 
similar neighbours? Do the two doublets of the AX system suddenly collapse to the singlet of the A2 
system? You have probably guessed that they do not. The transition is gradual. Suppose we have two 
different neighbours on an aromatic ring. The spectra show what we see. 





coupling is seen 
between 
these similar 
( neighbours 

two distorted 
doublets 



MeO 



L 



a 

The critical factor is how the 
difference between the chemical 
shifts of the two protons (A8) 
compares with the size of the 
coupling constant (/) for the 
machine in question. If A5 is 
much larger than / there is no 
distortion: if, say, A5 is 4 p.p.m. 
at 250 MHz (= 1000 Hz) and 
the coupling constant is a nor- 
mal 7 Hz, then this condition is 
fulfilled and we have an AX 
spectrum of two 1:1 doublets. 
As A8 approaches / in size, so 
the inner lines of the two dou- 
blets increase and the outer 
lines decrease until, when A5 is 
zero, the outer lines vanish away 
altogether and we are left with 
the two superimposed inner 
lines — a singlet or an A 2 spec- 
trum. You can see this progres- 
sion in the diagram. 



1 1 

2 p.p.m. 



A8» J 



AS> J 



A5~ J 



A5< J 



AX spectrum 



AM spectrum 



AB spectrum 



H A H B 



AB spectrum 



2H 



A5 = 



A 2 spectrum 



272 



11 ■ Proton nuclear magnetic resonance 



We call the last stages, where the distortion is great but the protons are still different, an AB spec- 
trum because you cannot really talk about H without also talking about H . The two inner lines 
may be closer than the gap between the doublets, or the four lines may all be equally spaced. Two 
versions of an AB spectrum are shown in the diagram — there are many more variations. 

It is a generally useful tip that a distorted doublet 'points' towards the protons with which it is coupled. 



You may see this situation 
described as an 'AB quartet'. It 
isn't! A quartet is an exactly 
equally spaced 1:3:3:1 system 
arising from coupling to three 
identical protons, and you should 
avoid this usage. 



doublets with a roof over their heads 




distorted 

doublet 

points 



proton coupled 
to this doublet 
is UPFIELD 



proton coupled 
to this doublet 
is DOWNFIELD 



X 



distorted 

doublet 

points 



Or, to put it another way, the AB system is 'roofed' with the usual arrangement of low walls and a 
high middle to the roof. Look out for doublets (or any other coupled signals) of this kind. 

We shall end this section with a final example illustrating para-disubstituted benzenes and roof- 
ing as well as an ABX system and an isopropyl group. 



h m H" CH-, 







_A_ 






p.p.m. 



The aromatic ring protons form a pair of distorted doublets (2H each) showing that the com- 
pound is a para-disubstituted benzene. Then the alkene protons form the AB part of an ABX spec- 
trum. They are coupled to each other with a large {trans) }= 16 Hz and one is also coupled to another 
distant proton. The large doublets are distorted (AB) but the small doublets within the right-hand half 
of the AB system are equal in height. The distant proton X is part of an f-Pr group and is coupled to 
H and the six identical methyl protons. Both /s are nearly the same so it is split by seven protons and 
is an octuplet. It looks like a sextuplet because the intensity ratios of the lines in an octuplet would be 
1:7:21:35:35:21:7:1 (from Pascal's triangle) and it is hardly surprising that the outside lines disappear. 

Coupling can occur between protons on the same carbon atom 

We have seen cases where protons on the same carbon atom are different: compounds with an alkene 
unsubstituted at one end. If these protons are different (and they are certainly near to each other), 
then they should couple. They do, but in this case the coupling constant is usually very small. Here 
you see the example we met on p. 000. 



1.4 Hj 




OSilVle. 




200 MHz 



SS 



JL 



p.p.m. 



Coupling in the proton NMR spectrum 



273 



The small 1.4 Hz coupling is a /hh coupling between two protons on the same carbon that are 
different because there is no rotation about the double bond, /hh coupling 
is called geminal coupling. 

This means that a monosubstituted alkene will have very characteristic sig- 
nals for the three protons on the double bond. The three different coupling 
constants are very different so that this ABX system is unusually clear. 

Here is an example of such a vinyl compound, ethyl acrylate (ethyl 
propenoate, a monomer for the formation of acrylic polymers). The spec- 
trum looks rather complex at first, but it is easy to sort out using the coupling 
constants. 



Proton NMR spectrum of a VINYL group 



Jab very small (0-2 Hz) 

J AX {cis) large (10-13 Hz) 

J BX {trans) very large (14-18 Hz) 






r 



H 3 C. Jb. 



200 MHz 



T 



1 p.pin. 



The largest / (16 Hz) is obviously between X and B {trans coupling), the medium / (10 Hz) is 
between X and A (cis coupling), and the small / (4 Hz) must be between A and B (geminal). This 
assigns all the protons: A, 5.80 p.p.m.; B, 6.40 p. p.m.; X, 6.11 p. p.m. Rather surprisingly, X comes 
between A and B in chemical shift. Assignments based on coupling are more reliable than those 
based on chemical shift alone. 

An enol ether type of vinyl group is present in ethyl vinyl ether, a reagent used for the protection 
of alcohols. This time all the coupling constants are smaller because of the electronegativity of the 
oxygen atom, which is now joined directly to the double bond. 



13 Hi* 
7H? 




H 3 C "0 

200 MHz in CDCI a 






1 p.p.m. 



It is still a simple matter to assign the protons of the vinyl group because couplings of 13, 7, and 2 
Hz must be trans, cis, and geminal, respectively. In addition, X is on a carbon atom next to oxygen 
and so goes downfield while A and B have extra shielding from the conjugation of the oxygen lone 
pairs (seep. 000). 

Geminal coupling on saturated carbons can be seen only if the hydrogens of a CH 2 group are 
different. We have seen an example of this on the bridging CH 2 group of myrtenal (p. 000). The 



2.49 H A 




274 11 ■ Proton nuclear magnetic resonance 



coupling constant for the protons on the bridge, 7ab> is 9 Hz. Geminal coupling constants in a satu- 
rated system can be much larger (typically 10-16 Hz) than in an unsaturated one. 



Typical coupling constants 


Geminal 2 J HH 


\/ HA 




saturated 


J|- 


10-16 Hz 




^=^ / h * 






|T%^ 




unsaturated 


1° 


0-3 Hz 


Vicinal 3 Jhh 


R^ R 




saturated 


1° 


6-8 Hz 




H A 






/k./ R 






R^^V^ 




unsaturated trans 


rl° 


14-16 Hz 




H* 

1 






r\ H 




unsaturated cis 


R 


8-11 Hz 




H* 






^r HB 




unsaturated aromatic 


O 1 


6-9 Hz 


Long-range 4 J H h 


H *-^^Y 




meta 


u 


1-3 Hz 




HA ^^r HB 




allylic 


R R 


1-2 Hz 





To conclude 

You have now met, in Chapter 3 and this chapter, all of the most important spectroscopic techniques 
available for working out the structure of organic molecules. We hope you can now appreciate that 
proton NMR is by far the most powerful of these techniques, and we hope you will be referring back 
to this chapter as you read the rest of the book. We shall talk about proton NMR a lot, and specifical- 
ly we will come back to it in detail in Chapter 15, where we will look at using all of the spectroscopic 
techniques in combination, and in Chapter 32, when we look at what NMR can tell us about the 
shape of molecules. 



Problems 



275 



Problems 

1. How many signals will there be in the H NMR spectrum of 5. Assign the H NMR spectra of these compounds and explain 

each of these compounds? Estimate the chemical shifts of the the multiplicity of the signals. 

signals. 



MeO. ,OMe 




Me 2 N 



X 






Me 



2. Comment on the chemical shifts of these three compounds 
and suggest whether there is a worthwhile correlation with pK a . 



Compound 


8 H , p.p.m. 


P«a 


CH 3 N0 2 


4.33 


10 


CH 2 (N0 2 )2 


6.10 


4 


CH(N0 2 ) 3 


7.52 






3. One isomer of dimethoxybenzoic acid has the H NMR spec- 
trum 3.85 (6H, s), 6.63 (1H, t, }2 Hz), 7.17 (2H, d, }2 Hz) and one 
isomer of coumalic acid has the H NMR spectrum 6.41 (1H, d, / 
10 Hz), 7.82 (1H, dd, }2, 10 Hz), 8.51 (1H, d, /2 Hz). In each case, 
which isomer is it? The substituents in black can be on any carbon 
atoms. 

.^X0 2 H 

L H0 2 C- .. 



MeO- 




OMe 



dimethoxybenzoic acid 



coumalic acid 



4. Assign the NMR spectra of this compound (assign means say 
which signal belongs to which atom) and justify your assign- 
ments. 




: . 



p.p.n 




5 H 0.97(3H,t,/7Hz), 
1.42(2H,sextuplet,/7Hz), 
2.00 (2H, quintet,/ 7 Hz), 
4.40(2H,t,/7Hz) 



5 H 1.08(6H,d,/7Hz), 
2.45(4H,t,/5Hz), 
2.80(4H,t,/5Hz), 
2.93(lH,septuplet,;7Hz) 




5Hl.00(3H,t,/7Hz), 
1.75(2H,sextuplet,/7Hz), 
2.91 (2H,t,/7Hz), 
7.4-7.9 (5H,m) 

6. The reaction below was expected to give product 6A and did 
indeed give a product with the correct molecular formula by mass 
spectrometry. The H NMR spectrum of the product was how- 
ever: 5 H (p.p.m.) 1.27 (6H, s), 1.70 (4H, m), 2.88 (2H, m), 5.4-6.1 
(2H, broad s, exchanges with D 2 0), 7.0-7.5 (3H, m). Though the 
detail is missing from this spectrum, how can you already tell that 
this is not the compound expected? 

HON ? .. ,P 





6A 

7. Assign the 400 MHz H NMR spectrum of this enynone as far 
as possible, justifying both chemical shifts and coupling patterns. 





TJ 




r 




■> — i - 



— 1 < 1 — 

3.0 1.0 



276 



11 ■ Proton nuclear magnetic resonance 



8. A nitration product (CgHn^C^) of this pyridine has been 
isolated which has a nitro (NO2) group somewhere on the mole- 
cule. From the 90 MHz H NMR spectrum, deduce whether the 
nitro group is (a) on the ring, (b) on the NH nitrogen atom, or (c) 
on the aliphatic side chain and then exactly where it is. Give a full 
analysis of the spectrum. 



[N0 2 somewhere 
in the molecule] 




JUiiL 




O || p: nr 



9. The natural product bullatenone was isolated in the 1950s 
from a New Zealand myrtle and assigned the structure 9A. Then 
compound 9A was synthesized and found not to be identical with 
natural bullatenone. Predict the expected H NMR spectrum of 
9A. Given the full spectroscopic data available nowadays, but not 
in the 1950s, say why 9A is definitely wrong and suggest a better 
structure for bullatenone. 

Spectra of bullatenone: 

Mass spectrum: m/z 188 (10%) (high resolu- 
tion confirms C^H^G^), 105 (20%), 102 
(100%), and 77 (20%) 
Ph" ^O" Infrared: 1604 and 1705 cnT 1 . 

„ 9* *H NMR: 1.45 (6H,s), 5.82 (lH,s), 7.35 (3H, 

alleged bullatenone ' 

m), and 7.68 (2H,m). 




10. Interpret this : H NMR spectrum. 

"x& JjLj*1 



;, 



L_1JL 



r 



11. Suggest structures for the products of these reactions, inter- 
preting the spectroscopic data. You are not expected to write 
mechanisms for the reactions and you should resist the tempta- 
tion to work out what 'should happen' from the reactions. These 
are all unexpected products. 




MeO v 



Br 



MeOH 



A, C5H12O2 
v ma ,(cm 4 ) 1745 
8 c (p.p.m.)179, 52, 39, 27 
8 H (p.p.m.) 1.20 (9H, s), 3.67 (3H, 



<^° 



EtOH 



B, C6H10O3 


Vmaxfcrrr 1 ) 1745, 1710 


8 c (p.p.m.) 203, 170, 62, 39, 22, 15 


8 H (p.p.m.) 1.28 (3H, t, J 7 Hz), 2.21 (3H, s), 


3.24 (2H, s), 4.2 (2H, q, 11 Hz) 



HO 




OMe 
SMe 



1. SOCI 2 , Et 3 N 



2. H 2 







C 






m 


/zll8 




V 


11 ax' 


cm" 1 ) 1730 




5c(P-P 


m. 


202, 45, 22, 


15 


S H (P 


p.n 


.) 1.12 (6H, s 




2.8 


3H 


s), 9.8 (1H, s 





12. Precocene is an compound that causes insect larvae to pupate 
and can also be found in some plants (Ageratum spp.) where it 
may act as an insecticide. It was isolated in minute amounts and 
has the following spectroscopic details. Propose a structure for 
precocene. 

Spectra of precocene: 

Mass spectrum: m/z (high resolution gives Ci3H 16 03), M — 15 

(100%) and M— 30 (weak). 

Infrared: CH and fingerprint only. 

J H NMR: 1.34 (6H, s), 3.80 (3H, s), 3.82 (3H, s), 5.54 (1H, d, / 10 

Hz), 6.37 (1H, d, /10 Hz), 6.42 (1H, s), and 6.58 (1H, s ). 

13. Suggest structures for the products of these reactions, inter- 
preting the spectroscopic data. Though these products, unlike 
those in Problem 11, are reasonably logical, you will not meet the 
mechanisms for the reactions until Chapters 22, 29, and 23, 
respectively, and you are advised to solve the structures through 
the spectra. 



OSiMe 3 



AICI 3 



*^N 



A, C 10 H 14 

v max( cml ) C ~ H ancl fingerprint only 

5 c (p.p.m.) 153, 141, 127, 115, 59, 33, 24 

S„(p.p.m.) 1.21 (6H, d, J 7 Hz), 2.83 (1H, septuplet, J 7 Hz), 

3.72 (3H, s), 5.74 (2H, d, J 9 Hz), and 7.18 (2h, d, J 9 Hz) 



C0 2 Me 





B, C8H14O3 








v„„(cm 4 ) 1745, 1730 






5 c (p.p.m 


) 202, 176, 62, 48, 34 


22 


15 


5 H (p.p.m 


) 1.21 (6H,s), 1.8{2H,t 


J 7 Hz), 


2.24 (2H,t 


J 7 Hz), 4.3 (3H, s), 10.01 


(1H.S) 



Me 2 N' 





C, CiiHi 


5 N0 2 








v mas (cm' 1 


1730 






5 c (p.p.m 


) 191, 164, 132 


130, 115 


64, 41, 


29 


5 H (p.p 


m.)2.32(6H, s), 


3.05 (2H, 


, J 6 Hz 




4.20 


(2H, t, J6 Hz), 6.97 (2H, d, 


J 7 Hz), 




7.82 (2H, d, J 7 H 


I), 9.97 (11- 


,s) 





Problems 



277 



14. The following reaction between a phosphonium salt, base, 
and an aldehyde gives a hydrocarbon CsH^ with the 200 MHz H 
NMR spectrum shown. Give a structure for the product and com- 
ment on its stereochemistry. You are not expected to discuss the 
chemistry! 



pii,p v 



1. NaNH 



r 



^dlLrfiL £ 



it 

JUL 



^JK^_ 



1 



L_ 



G P.Z'.iTi. 



Nucleophilic substitution at the 
carbonyl (C=0) group 



12 



Connections 

Building on: 

• Drawing mechanisms ch5 

• Nucleophilic attack on carbonyl 
groups ch6 & ch9 

• Acidity and pK a ch8 

• Grignard and RLi addition to C=0 
groups ch9 



Arriving at: 

• Nucleophilic attack followed by loss of 
leaving group 

• What makes a good nucleophile 

• What makes a good leaving group 

• There is always a tetrahedral 
intermediate 

• How to make acid derivatives 

• Reactivity of acid derivatives 

• How to make ketones from acids 

• How to reduce acids to alcohols 



You are already familiar with reactions of compounds containing carbonyl groups. Aldehydes and 
ketones react with nucleophiles at the carbon atom of their carbonyl group to give products contain- 
ing hydroxyl groups. Because the carbonyl group is such a good electrophile, it reacts with a wide 
range of different nucleophiles: you have met reactions of aldehydes and ketones with (in Chapter 6) 
cyanide, water, alcohols, and (in Chapter 9) organometallic reagents (organolithiums and organo- 
magnesiums, or Grignard reagents). 

In this chapter and Chapter 14 we shall look at some more reactions of the carbonyl group — and 
revisit some of the ones we touched on in Chapter 6. It is a tribute to the importance of this function- 
al group in organic chemistry that we have devoted four chapters of this book to its reactions. Just 
like the reactions in Chapters 6 and 9, the reactions in Chapters 12 and 14 all involve attack of a 
nucleophile on a carbonyl group. The difference will be that this step is followed by other mechanis- 
tic steps, which means that the overall reactions are not just additionsbut also substitutions. 

The product of nucleophilic addition to a carbonyl group is 
not always a stable compound 

Addition of a Grignard reagent to an aldehyde or ketone gives a stable alkoxide, which can be proto- 
nated with acid to produce an alcohol (you met this reaction in Chapter 9). 

The same is not true for addition of an alcohol to a carbonyl group in the presence of base — in 
Chapter 6 we drew a reversible, equilibrium arrow for this transformation and said that the product, 
a hemiacetal, is only formed to a significant extent if it is cyclic. 

The reason for this instability is that RO~ is easily expelled from the molecule. We call groups that 
can be expelled from molecules, usually taking with them a negative charge, leaving groups. We'll 
look at leaving groups in more detail later in this chapter and again in Chapter 17. 



Looking forward to: 

• Loss of carbonyl oxygen chl4 

• Kinetics and mechanism chl3 

• Reactions of enols ch21, ch26-ch29 

• Synthesis in action ch25 



JL 



R^H^" - 



,/ v 



OR 



(^ 



unstable 
intermediate 



u 

A 




OR 



1. EtMgBr 

2. H 3 + HO. 



* ^* "^ 



ROH HO OR 



^ ^T^ 



ketone 



hemiacetal 



RO is a leaving group 



280 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



Leaving groups 

Leaving groups are anions such as Cl~, RO~, and RCO2 that can be expelled from 
molecules taking their negative charge with them. 



So, if the nucleophile is also a leaving group, there is a chance that it will be lost again and that the 
carbonyl group will reform — in other words, the reaction will be reversible. The energy released in 
forming the C=0 bond (bond strength 720 kj mol - ) more than makes up for the loss of two C-O 
single bonds (about 350 kJmol~ each), one of the reasons for the instability of the hemiacetal prod- 
uct in this case. 



51 

Me'lr'OR 

VJf e 

MgBr 



9? 



Me 



Me 



unstable 
intermediate 




®OR I RO ' s a leaving group 



Me Me 



ketone 
reacts 
further 



Again, it collapses with loss of RO~ as a leaving group. This time, though, we have not gone back 
to starting materials: instead we have made a new compound (a ketone) by a substitution reaction — 
the OR group of the starting material has been substituted by the Me group of the product. In fact, as 
we shall see later, this reaction does not stop at this point because the ketone product can react with 
the Grignard reagent a second time. 






R ^OH 

carboxylic acid 



R^X 

carboxylic acid derivatives 



The reactions of alcohols with 
acid chlorides and with acid 
anhydrides are the most 
important ways of making esters, 
but not the only ways. We shall 
see later how carboxylic acids can 
be made to react directly with 
alcohols. Rememberthe 
convenient organic element 
symbol for 'acetyl', Ac? Cyclohexyl 
acetate can be represented by 
'OAc' but not just 'Ac'. 



cyclohexyl 
acetate can 
be drawn 
like this: 



OAc 



But NOT 
like this: 





Carboxylic acid derivatives 



Most of the starting materials for, and products of, these substitutions will be carboxylic acid deriva- 
tives, with the general formula RCOX. You met the most important members of this class in Chapter 
3: here they are again as a reminder. 



Carboxylic acid derivatives 





A 



R ^Cl 






acid chlorides (acyl chlorides) 3 



acid anhydrides 



x 

R^A)" 1 


A. 



R 0R esters 



x 



2 amides 



a We shall use these two terms interchangeably. 



Acid chlorides and acid anhydrides react with alcohols to make esters 

Acetyl chloride will react with an alcohol in the presence of a base to give an acetate ester and we get 
the same product if we use acetic anhydride. 




ci^A 

acetyl chloride 



cyclohexanol 

» 

base 





v OH 

cyclohexanol 



oX~ 



base 



cyclohexyl acetate 



O U 

AA 



acetic anhydride 



In each case, a substitution (of the black part of the molecule, CI or AcO , by the orange 
cyclohexanol) has taken place — but how? It is important that you learn not only the fact that 



Carboxylic acid derivatives 



281 



You will notice that the terms 
'acid chloride' and 'acyl chloride' 
are used interchangeably. 



acyl chlorides and acid anhydrides react with alcohols but also the mechanism of the reaction. In this 
chapter you will meet a lot of reactions, but relatively few mechanisms — once you understand one, 
you should find that the rest follow on quite logically. 

The first step of the reaction is, q-^. 

as you might expect, addition of 11^ 

the nucleophilic alcohol to the Cl^*^ 

electrophilic carbonyl group — 
we'll take the acyl chloride first. ^^ ^QH 

The base is important because 
it removes the proton from the 
alcohol as it attacks the carbonyl 





base 



group. A base commonly used for this is pyridine. If the electrophile had been an aldehyde or a 

ketone, we would have got an unstable hemiacetal, which would collapse back to starting materials 

by eliminating the alcohol. With an acyl chloride, the alkoxide intermediate we get is also unstable. It 

collapses again by an elimination 

leaving ~\ 

group \^J 



cr 



rr 



L 



unstable 
^ — intermediate 



-0, 

CI" is a 
leaving group 




oA 



reaction, this time losing chloride 
ion, and forming the ester. 
Chloride is the leaving group 
here — it leaves with its negative 
charge. 

With this reaction as a model, 
you should be able to work out 
the mechanism of ester formation from acetic anhydride and cyclohexanol. Try to write it down 
without looking at the acyl chloride mechanism above, and certainly not at the answer below. Here it 
is, with pyridine as the base. Again, addition of the nucleophile gives an unstable intermediate, which 
undergoes an elimination reaction, this time losing a carboxylate anion, to give an ester. 

anhydride starting material 



Cf) 

A ^ 




alcohol starting material 




pyridine 



We call the unstable intermediate formed in these reactions the tetrahedral intermediate, be- 
cause the trigonal (sp ) carbon atom of the carbonyl group has become a tetrahedral (sp ) carbon 
atom. 



Tetrahedral intermediates 

Substitutions at trigonal carbonyl groups go through a tetrahedral intermediate 
and then on to a trigonal product. 




Nu R 



£ 



trigonal planar 
starting material 



Q 



Nu OS 

R^X 

tetrahedral 
intermediate 



u 

Nu^N 



trigonal planar 
product 



282 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



More details of this reaction 

This reaction has more subtleties than first meet 
the eye. If you are reading this chapter for the first 
time, you should skip this box, as it is not essential 
to the general flow of what we are saying. There are 
three more points to notice. 

1 Pyridine is consumed during both of these 
reactions, since it ends up protonated. One 
whole equivalent of pyridine is therefore 
necessary and, in fact, the reactions are often 
carried out with pyridine as solvent 

Nucleophilic catalysis in ester formation 



^ 





©N 




2 The observant among you may also have 

noticed that the (weak — pyridine) base catalyst 
in this reaction works very slightly differently 
from the (strong — hydroxide) base catalyst 
in the hemiacetal-forming reaction on p. 000: 
one removes the proton after the nucleophile 
has added ; the other removes the proton 
before the nucleophile has added. This is 
deliberate, and will be discussed further in 
Chapter 13 



-H + 




tetrahedral 
intermediate 



ROH 

reactive trigonal 
intermediate 



3 Pyridine is, in fact, more nucleophilic than the 
alcohol, and it attacks the acyl chloride rapidly, 
forming a highly electrophilic (because of the 
positive charge) intermediate. It is then this 
intermediate that subsequently reacts with the 
alcohol to give the ester. Because pyridine is 
acting as a nucleophile to speed up the reaction, 
yet is unchanged by the reaction, it is called a 
nucleophilic catalyst. 




u 

A 0R + 1 




tetrahedral 
intermediate 



Non-radioactive isotopes are 
detected by mass spectrometry 
(Chapter 3). 



How do we know that the tetrahedral intermediate exists? 

We don't expect you to be satisfied with the bland statement that tetrahedral intermediates are 
formed in these reactions: of course, you wonder how we know that this is true. The first evidence for 
tetrahedral intermediates in the substitu- 
tion reactions of carboxylic acid deriva- i 8 _ is n 



tives was provided by Bender in 1951. He 
reacted water with carboxylic acid deriva- 
tives RCOX that had been 'labelled' with 



rapid migration of 

the hydrogen atom 

between the 

oxygen atoms 



18, 



plus 



OH 





18 0H 



an isotope of oxygen, O. 

He then reacted these derivatives with water to make labelled carboxylic acids. However, he added 
insufficient water for complete consumption of the starting material. At the end of the reaction, he found 
that the proportion of labelled molecules in the remaining starting materialhad decreased significantly: in 
other words, it was no longer completely labelled with O; some contained 'normal' O. 

This result cannot be explained by direct substitution of X by H2O, but is consistent with the exis- 
tence of an intermediate in which the unlabelled O and labelled O can 'change places'. This 
intermediate is the tetrahedral intermediate fox this reaction. 

18 © rapid migration of l&§\i 

protons between V n 2 



tetrahedral intermediate 




but it can also revert to unlabelled starting materia 
g>H 2 

R o x ~~ rAx 



Why are the tetrahedral intermediates unstable? 

The alkoxide formed by addition of a Grignard reagent to an aldehyde or ketone is stable. 
Tetrahedral intermediates are similarly formed by addition of a nucleophile to a carbonyl group, so 
why are they unstable 7 . The answer is to do with leaving group ability. 



Carboxylic acid derivatives 



283 



Once the nucleophile has added to the carbonyl compound, the stability of the product (or tetra- 
hedral intermediate) depends on how good the groups attached to the new tetrahedral carbon atom 
are at leaving with the negative charge. In order for the tetrahedral intermediate to collapse (and 
therefore be just an intermediate and not the final product) one of the groups has to be able to leave 
and carry off the negative charge from the alkoxide anion formed in the addition. 



CI 






EtOH 



base 



Cr / ^OEt 
Me 



Here once again is the tetrahedral intermediate resulting n n O 

from addition of an alcohol to an acyl chloride. 

There are three choices of leaving group: CI - , EtO~, and 
Me - . We cannot actually make Me~ because it is so un- 
stable, but MeLi, which is about as close to it as we can get (Chapter 9), reacts vigorously with water 
so Me~ must be a very bad leaving group. EtO~ is not so bad — alkoxide salts are stable, but they are 
still strong, reactive bases (we shall see below what pJC a has to do with this matter). But CI" is the best 
leaving group: CI" ions are perfectly stable and quite unreactive and happily carry off the negative 
charge from the oxygen atom. You probably eat several grams of CI" every day but you would be 
unwise to eat EtO" or MeLi. 

piC a H is a useful guide to leaving group ability 

It's useful to be able to compare leaving group ability quantitatively. This is impossible to do exactly, 
but a good guide is p-K" a H- If we go back to the example of ester formation from acyl chloride plus 
alcohol, there's a choice of Me - , EtO - , and CI". The leaving group with the lowest pK a n is the best 
and so we can complete the reaction. 



Leaving group 


P«aH 


Me" 


50 


Etcr 


16 


cr 


-7 






EtOH 



CK Me 



base 



CK/^OEt Me^^ 



Me 



OEt 



Remember that we use the term pK aH 
to mean 'pK a ofthe conjugate acid': if 
you need reminding about pK a and 
pK a n, stop now and refresh your 
memory by reviewing Chapter 8. 



The same is true for the reaction of acetic anhydride with an alcohol. Possible leaving groups from 
this tetrahedral intermediate are the following. 



Leaving group 


P«aH 


Me" 


50 


RO" 


16 


MeCO^ 


5 



X X 



ROH 



base 



Me 



e. 




„© 



OR 



Me 



AcO 

I 

Me' '"OR 






Again the group that leaves is the one with the lowest P-fC a H. 

• Leaving group ability 

The lower the pK aH , the better the leaving group in carbonyl substitution 
reactions. 

Why should this be so? The ability of an anion to behave as a leaving group depends in 
some way on its stability — how willing it is to accept a negative charge. pK a represents the 
equilibrium between an acid and its conjugate base, and is a measure of the stability of 
that conjugate base with respect to the acid — low pK a means stable conjugate base, indi- 
cating a willingness to accept a negative charge. So the general trends that affect pK a , 
which we discussed in Chapter 8, will also affect leaving group ability. However, you 
must bear in mind that pK a is a measure of stability only with respect to the protonated form 
of the anion. Leaving group ability is a fundamentally different comparison between the 
stability of the negatively charged tetrahedral intermediate and the leaving group plus result- 
ing carbonyl compound. But it still works as a good guide. These five values are worth 
learning. 



Leaving group 


P«aH 


R" 


50 


NHi 


35 


RO" 


16 


RCO2 


5 


cr 


-7 



284 12 ■ Nucleophilic substitution at the carbonyl (C=0) group 

We can use p-K" a to predict what happens F q 0® 

if we react an acyl chloride with a carboxylate ^ 

salt. We expect the carboxylate salt (here, sodium ci^ s Me CI / H 

formate, or sodium methanoate, HCC^Na) to ( 

act as the nucleophile to form a tetrahedral inter- ^0\ ^ 

mediate, which could collapse in any one of three 
ways. Na 

We can straight away rule out loss of Me~ (p-KaH 50), but we might guess that Cl~ (p^ a H _ 7) is a 
better leaving group than HCO2 (p^a about 5), and we'd be right. Sodium formate reacts with acetyl 
chloride to give 'acetic formic anhydride'. 

HC0 2 Na 



A 23 'C, 6 h II II 



Cr Me *- Me 



mixed anhydride 
64% yield 



r<> \ 

S^ M « ^X - M .A„A H 



Cr_ Me 

>0 H 

Na® I 



Amines react with acyl chlorides to give amides 

Using the principles we've outlined above, you should be able to see how these compounds can be 
interconverted by substitution reactions with appropriate nucleophiles. We've seen that acid 
chlorides react with carboxylic acids to 
give acid anhydrides, and with alcohols NH 3 





Me Me 



to give esters. They'll also react with ^r"^ ^Cl ^r NH 2 

° ' H 2 0, °C, 1 h 

amines (such as ammonia) to give 
amides. 

The mechanism is very similar to the 
mechanism of ester formation. 



Of e ? <3 








C'- JM 



Me^ ^X. Me v A^ Me. A^T* Me ^ 

S CI *~ N^V^CI *~ >" "NH 2 




NH 



2 




Me s -„ Me/ H^ Me Me 

NH 3 H H ) ©0 

:NH 3 »- NH4 CI 

Notice the second molecule of ammonia, which removes a proton before the loss of chloride 
ion — the leaving group — to form the amide. Ammonium chloride is formed as a by-product in the 
reaction. 

Here is another example, using a secondary amine, dimethylamine. Try writing down the mecha- 
nism now without looking at the one above. Again, two equivalents of dimethylamine are necessary, 
though the chemists who published this reaction added three for good measure. 




Me 2 NH 



0°C, 2 h 



aV 



(3equiv.) ^^ II Me @ 

^N + Me 2 NH 2 Cl u 

Me 



86-89% yield 



Carboxylic acid derivatives 



285 



Schotten-Baumann synthesis of an amide 



As these mechanisms show, the formation of 
amides from acid chlorides and amines is 
accompanied by production of one equivalent of 
HCI, which needs to be neutralized by a second 
equivalent of amine. An alternative method for 
making amides is to carry out the reaction in the 
presence of another base, such as NaOH, which 
then does the job of neutralizing the HCI. The 
trouble is, OhT also attacks acyl chlorides to 



give carboxylic acids. Schotten and Baumann, in 
the late nineteenth century, published a way round 
this problem by carrying out these reactions in 
two-phase systems of immiscible water and 
dichloromethane. (Carl Schotten (1853-1910) 
was Hofmann's assistant in Berlin and spent most 
of his working life in the German patent office. 
(There is more about Hofmann in Chapter 19.) The 
organic amine (not necessarily ammonia) and the 



Schotten-Baumann synthesis of an amide 




NaOH 
H 2 0, CH 2 CI 2 





acyl chloride remain in the (lower) dichloromethane 
layer, while the base (NaOH) remains in the (upper) 
aqueous layer. Dichloromethane and chloroform 
are two common organic solvents that are heavier 
(more dense) than water. The acyl chloride reacts 
only with the amine, but the HCI produced can 
dissolve in, and be neutralized by, the aqueous 
solution of NaOH. 



upper layer: 
aqueous solution 
of NaOH 



80% yield 



lower layer: 
dichloromethane 
solution of amine 
and acid chloride 



Using piC a H to predict the outcome of substitution reactions of carboxylic acid 
derivatives 




NH 3 



OMe 



You saw that acid anhydrides react with 
alcohols to give esters: they will also 
react with amines to give amides. But 
would you expect esters to react with 
amines to give amides, or amides to 
react with alcohols to give esters? Both 
appear reasonable. 

In fact only the top reaction works: amides can be formed from esters but esters cannot be formed 
from amides. Again, looking at pi^s can tell us why. In both cases, the tetrahedral intermediate 
would be the same. The possible leaving groups are shown in the table. 



MeOH 





OMe 



NH 3 ? 





O 
I 


MeOH? 


^Y 


\*OMe 








NH 2 




KJ 






tetrahedral 


intermediate 






So RO~ leaves and the amide is formed. Here is an example. The base may be either the EtCT pro- 
duced in the previous step or another molecule of PI1NH2. 

PhNH 2 

x 1 - 11 

h^^^^^-oct 135 C 1 h Ph-^ \^ ^ 



Ph' 



^OEt 



'NHPh 




OEt 



Ph 



OEt 



0°) 



Ph' 



base 






Ph 



PhHN 



O 

V\ 

H H 

base \^ 



O O 

X X 



Possible leaving groups 


P«aH 


Ph" 


45 


NH2 


35 


MeCT 


16 



You will meet many more 
mechanisms like this, in which an 
unspecified base removes a 
proton from an intermediate. As 
long as you can satisfy yourself 
that there is a base available to 
perform the task, it is quite 
acceptable to write either of 
these shorthand mechanisms. 



00 00 00 

XX JUr XX 






286 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



no 
reaction 




You saw pyridine doing this on p. 
000 — it's called general base 
catalysis, and we will talk about it in 
more detail in Chapter 13. 



Factors other than leaving group ability can be important 

In fact, the tetrahedral intermediate would simply never form from an amide and an alcohol; the 
amide is too bad an electrophile and the alcohol not a good enough nucleophile. We've looked at 
leaving group ability: next we'll consider the strength of the nucleophile Y and then the strength of 
the electrophile RCOX. 



Conditions for reaction 

If this reaction is to go 

o 

A. 



,© 



► .A. 



,e 



1. X must be a better leaving group than Y (otherwise the reverse reaction 
would take place) 

2. Y must be a strong enough nucleophile to attack RCOX 

3. RCOX must be a good enough electrophile to react with Y~ 



piC a H is a guide to nucleophilicity 

We have seen how pK a gives us a guide to leaving group ability: it is also a good guide to how strong 
a nucleophile will be. These two properties are the reverse of each other: good nucleophiles are bad 
leaving groups. A species that likes forming new bonds to hydrogen (in other words, the pK a of its 
conjugate acid is high) will also like to form new bonds to carbon: it is likely to be a good nucleo- 
phile. Bases with high p-KaH are bad leaving groups and they are, in general, good nucleophiles 
towards the carbonyl group. We will come back to this concept again in Chapter 17, where you will 
see that it does not apply to substitution at saturated carbon atoms. 

• Guide to nucleophilicity 

In general, the higher the piC aH , the better the nucleophile. 

But just a moment — we've overlooked an important point. 
When we made acid anhydrides from acid chlorides plus car- 
boxylate salts, we used an anionic nucleophile RCO2 but, when 
we made amides from acid chlorides plus amines, we used a neu- 
tral nucleophile NH3, and not NH2. For proper comparisons, we 
should include in our table ROH (piCjH = _ 5; in other words, -5 
is the pK a of ROHj ) and NH 3 (pJC aH = 9; in other words, 9 is the 
piC a ofNHj). 

While amines react with acetic anhydride quite rapidly at 
room temperature (reaction complete in a few hours), alcohols 
react extremely slowly in the absence of a base. On the other 
hand, an alkoxide anion reacts with acetic anhydride extremely 

rapidly — the reactions are often complete within seconds at °C. We don't have to deprotonate an 
alcohol completely to increase its reactivity: just a catalytic quantity of a weak base can do this job by 
removing the alcohol's proton as it adds to the carbonyl group. All these observations are consistent 
with our table and our proposition that high pX" a H means good nucleophilicity. 

Not all carboxylic acid derivatives are equally reactive 

We can list the common carboxylic acid derivatives in a 'hierarchy' of reactivity, with the most reac- 
tive at the top and the least reactive at the bottom. Transformations are always possible moving down 



I 


1 Base 


P«aH 




FT 


50 


I 

Q- 
01 

£Z 

\n 

CD 
CD 


NH2 
R0 _ 

NH 3 


35 
16 
9 


£Z 


RCO2 


5 




ROH 


-5 




cr 


-7 



Not all carboxylic acid derivatives are equally reactrive 



287 



the hierarchy. We've seen that this hierarchy is partly due to how good the leaving group is (the ones 
at the top are best), and partly due to how good the nucleophile needed to make the derivative is (the 
ones at the bottom are best). 



most reactive 



II acid chlorides 

R^^CI (acyl chlorides) 





R^O^R* 


acid anhydrides 




u 

R-'^OR 1 


esters 




x 


amides 



least reactive 



Derealization and the electrophilicity of carbonyl compounds 

All of these derivatives will react with water to form carboxylic acids, but at very different rates. 

JL. ~^ JL U fastat2 °" c JL- -— - JL 



R' ^Cl 


AA 



H 2 



x 



slow at 20 'C 



A 



R OH 

H 2 



only on heating with 
acid or base catalyst 



NH 2 



»^N)H 



only on prolonged heating 
with strong acid 
or base catalyst 



Hydrolysing an amide requires boiling in 10% NaOH or heating overnight in a sealed tube with 
concentrated HC1. Amides are the least reactive towards nucleophiles because they exhibit the great- 
est degree of delocalization. You met this concept in Chapter 7 and we shall return to it many times 
more. In an amide, the lone pair on the nitrogen atom can be stabilized by overlap with the 7t* orbital 
of the carbonyl group — this overlap is best when the lone pair occupies a p orbital (in an amine, it 
would occupy an sp orbital). 



K K 



o e 



molecular orbital diagram shows how energy of 
orbitals changes as lone pair and C=0 Jt* interact 

, , new higher-energy jt* orbital 



lone pair in 
p orbital 



orbitals overlap 




empty n* 
orbital 



isolated 
lone pa 

on 



„r -44-t 



allow orbitals to 
interact 



MP 



isolated C=0 
jt* orbital 



new, stabilized lower-energy lone pair 



The molecular orbital diagram shows how this interaction both lowers the energy of the bonding 
orbital (the delocalized nitrogen lone pair), making it neither basic nor nucleophilic, and raises the 
energy of the 71* orbital, making it less ready to react with nucleophiles. Esters are similar but, 
because the oxygen lone pairs are lower in energy, the effect is less pronounced. 

The greater the degree of delocalization, the weaker the C=0 bond becomes. This is most clearly 



288 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



evident in the stretching frequency of the carbonyl group in the IR spectra of carboxylic acid deriva- 
tives — remember that the stretching frequency depends on the force constant of the bond, itself a 
measure of the bond's strength (the carboxylate anion is included because it represents the limit of 
the series, with complete derealization of the negative charge over the two oxygen atoms). 


JL. JUL JL. JL. JL 



We treat this in more detail in Chapter 
15. There are two frequencies for the 
anhydride and the carboxylate because 
of symmetric and antisymmetric 
stretching. 



V / cm" 1 



C=0 



XI 



1790-1815 



1800-1850 
1740-1790 



'OR 



1735-1750 



NH 2 



1690 



strongest 



0© 

1610-1650 
1300-1420 

»- weakest 



Amides react as electrophiles only with powerful nucleophiles such as HCT. Acid chlorides, on the 
other hand, react with even quite weak nucleophiles: neutral ROH, for example. They are more reac- 
tive because the electron-withdrawing effect of the chlorine atom increases the electrophilicity of the 
carbonyl carbon atom. 



Bond strengths and reactivity 

You may think that a weaker C=0 bond should be 
more reactive. This is not so because the partial 
positive charge on carbon is also lessened by 
derealization and because the molecule as a 
whole is stabilized by the derealization. Bond 
strength is not always a good guide to reactivity! 

For example, in acetic acid the bond strengths are 
surprising. The strongest bond is the 0-H bond and 
the weakest is the C-C bond. Yet very few reactions 



of acetic acid involve breaking the C-C bond, and its 
characteristic reactivity, as an acid, involves 
breaking 0-H, the strongest bond of them all! 

The reason is that polarization of bonds and 
solvation of ions play an enormously important role 
in determining the reactivity of molecules. In 
Chapter 39 you will see that radicals are relatively 
unaffected by solvation and that their reactions 
follow bond strengths much more closely. 



bond energies in kJ mol 1 41 g 



456 



469 



-0 






351 (a) 
+369 (it) 



339 



Later in this chapter (p. 000) you will 
meet about the only nucleophiles that 
will: organolithium compounds attack 
lithium carboxylates. 



Carboxylic acids do not undergo substitution reactions under basic conditions 

Substitution reactions of RCO2H require a leaving group OH~, with P-K" a H = 15, so we should be able 
to slot RCO2H into the 'hierarchy' on p. 000 just above the esters RCC^R'- However, if we try to 
react carboxylic acids with alcohols in the presence of a base (as we would to make esters from acyl 
chlorides), the only thing that happens is deprotonation of the acid to give the carboxylate anion. 
Similarly, carboxylic acids react with amines to give not amides but ammonium carboxylate salts, 
because the amines themselves are basic. 






amide 
not formed 



NH 3 , 20 °C 



NH 3 , 20 °C 



OH 



NH 



© 







ammonium salt 
(ammonium acetate) 



Once the carboxylic acid is deprotonated, substitutions are prevented because (almost) no 
nucleophile will attack the carboxylate anion. Under neutral conditions, alcohols are just not reac- 
tive enough to add to the carboxylic acid but, with acid catalysis, esters can be formed from alcohols 
and carboxylic acids. 



In fact, amides can be made from carboxylic acids plus 
amines, but only if the ammonium salt is heated stronglyto 
dehydrate it. This is not usually a good way of making 
amides! 



140-210 C 

0© ' NH 2 






NH 4 



+ H 2 



Acid catalysts increase the reactivity of a carbonyl group 

We saw in Chapter 6 that the lone pairs of a carbonyl group may be proto- 
nated by acid. Only strong acids are powerful enough to protonate carbonyl 
groups: the pK a of protonated acetone is -7, so, for example, even 1M HC1 
(pH 0) would protonate only 1 in 10 molecules of acetone. However, even 
proportions as low as this are sufficient to increase the rate of substitution 
reactions at carbonyl groups enormously, because those carbonyl groups that 
are protonated become extremely powerful electrophiles. 



Not all carboxylic acid derivatives are equally reactive 



289 






0/H 



-A, 



protonated carbonyl group is 
a powerful electrophile 



It is for this reason that alcohols will react with carboxylic acids under acid catalysis. The acid 
(usually HC1, or H2SO4) reversibly protonates a small percentage of the carboxylic acid molecules, 
and the protonated carboxylic acids are extremely susceptible to attack by even a weak nucleophile 
such as an alcohol. 

acid-catalysed ester formation: forming the tetrahedral intermediate 

■• .£> ^ H 

f*V HO OH HO OH 

^^ ^ R ^^ X - R 



u 

^OH 



starting material 



'? 



tetrahedral 
intermediate 



Acid catalysts can make bad leaving groups into good ones 

This tetrahedral intermediate is unstable because the energy to be gained by re-forming a C=0 bond 
is greater than that used in breaking two C-O bonds. As it stands, none of the leaving groups (R~, 
HO~, or RO~) is very good. However, help is again at hand in the acid catalyst. It can protonate any 
of the oxygen atoms reversibly. Again, only a very small proportion of molecules are protonated at 
any one time but, once the oxygen atom of, say, one of the OH groups is protonated, it becomes a 
much better leaving group (H 2 0, P-K" a H _ 2, instead of HO - , pK a n 15). Loss of ROH from the tetra- 
hedral intermediate is also possible: this leads back to starting materials — hence the equilibrium 
arrow in the scheme above. Loss of H2O is more fruitful, and takes the reaction forwards to the ester 
product. 
acid-catalysed ester formation (continued) 



.Oh® 



HO. 




®o2 

-A 



tetrahedral 
intermediate 



H 2 



A .» 

ester 
product 



Acid catalysts catalyse substitution reactions of carboxylic acids 

1 They increase the electrophilicity of the carbonyl group by protonation at 
carbonyl oxygen 

2 They lower the pK a n of the leaving group by protonation there too 



Average bond strength C=0 = 
720 kJ mol -1 . Average bond 
strength C-0 = 351 kJ mol" 1 



We shall discuss the reasons why 
chemists believe this to be the 
mechanism of this reaction later in the 
chapter. 



Ester formation is reversible: how to control an equilibrium 

Loss of water from the tetrahedral intermediate is reversible too: just as ROH will attack a protonat- 
ed carboxylic acid, H2O will attack a protonated ester. In fact, every step in the sequence from car- 
boxylic acid to ester is an equilibrium, and the overall equilibrium constant is about 1. In order for 
this reaction to be useful, it is therefore necessary to ensure that the equilibrium is pushed towards 
the ester side by using an excess of alcohol or carboxylic acid (usually the reactions are done in a 
solution of the alcohol or the carboxylic acid). In this reaction, for example, using less than three 
equivalents of ethanol gave lower yields of ester. 
3 equiv. EtOH 



R0' 



X0 2 H 



dry HCI gas 



RO^ X0 2 Et 

68-72% yield 



290 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



Alternatively, the reaction can be done in the presence of a dehydrating agent (concentrated 
H2SO4, for example, or silica gel), or the water can be distilled out of the mixture as it forms. 



Lactic acid (the 
structure is shown in an 
example on this page) 
must be handled in 
solution in water. Can 
you see why, bearing in 
mind what we have said 
about the reversibility of 
ester formation? 




OH 




OH 

lactic acid 



cat. H 2 S0 4 
benzene (solvent) OH 

remove water by distillation 89-91% yield 



OH 



AcOH 



cat. H 2 S0 4 

silica gel 

(drying agent) 



V 



57% yield 



You have now met three ways of making esters from alcohols: 

• with acyl chlorides • with acid anhydrides • with carboxylic acids 

Try to appreciate that different methods will be appropriate at different times. If you 

want to make a few milligrams of a complex ester, you are much more likely to work 

with a reactive acyl chloride or anhydride, using pyridine as a weakly basic catalyst, 

than to try and distil out a minute quantity of water from a reaction mixture containing a 

strong acid that may destroy the starting material. On the other hand, if you are a 

chemist making simple esters (such as those in Chapter 3, p. 000) for the flavouring 

industry on a scale of many tons, you will prefer the cheaper option of carboxylic acid 

plus HCI in alcohol solution. 



Acid-catalysed ester hydrolysis and 
transesterification 

By starting with an ester, an excess of water, and an 
acid catalyst, we can persuade the reverse reaction to 
occur: formation of the carboxylic acid plus alcohol 
with consumption of water. Such a reaction is known 
as a hydrolysis reaction, because water is used to break 
up the ester into carboxylic acid plus alcohol (lysis = 
breaking). 



excess water forces 
reaction forward 



acid-catalysed ester hydrolysis 



acid-catalysed ester formation 



excess ester or removal of water forces 
the reaction backward 



.© 



y 

A — 5L 



J 



H 2 



The mechanisms of acid-catalysed 
formation and hydrolysis of esters 
are extremely important: you must 
learn them, and understand the 
reason for each step. 



HO OR 



HO. OR 



^OH 






© 



W 



OH 
ROH 



OH 



Acid-catalysed ester formation and hydrolysis are the exact reverse of one another: the only way 
we can control the reaction is by altering concentrations of reagents to drive the reaction the way we 
want it to go. The same principles can be used to convert to convert an ester of one alcohol into an 
ester of another, a process known as transesterification. It is possible, for example, to force this equi- 
librium to the right by distilling methanol (which has a lower boiling point than the other compo- 
nents of the reaction) out of the mixture. 



OMe 



cat. HCI 



MeOH 





The mechanism for this transesterification simply consists of adding one alcohol (here BuOH) 
and eliminating the other (here MeOH), both processes being acid-catalysed. Notice how easy it is 
now to confirm that the reaction is catalytic in H + . Notice also that protonation always occurs on the 
carbonyl oxygen atom. 



OMe 



H© 



OBu 





94% yield 



^^JoMe 

VOH 



© 



OBu 




irreversible because MeOH 
is removed from the mixture 



®,0 



(MeOH] \ -* 



distilled off 



Not all carboxylic acid derivatives are equally reactive 



291 



Polyester fibre manufacture 

A transesterification reaction is used to make the 
polyester fibres that are used for textile production. 
Terylene, or Dacron, for example, is a polyester of the 



dicarboxylic acid terephthalic acid and the diol ethylene 
glycol. Polymers are discussed in more detail in Chapter 
52. 




terephthalic acid 



ethylene glycol 



Dacron® or Terylene - a polyester fibre 



It is made by transesterifying diethyl 
terephthalatewith ethylene glycol in 
the presence of an acid catalyst, 
distilling off the methanol as it forms. 



MeO 




OMe 



Dacron® or Terylene 



cat. H + 



Base-catalysed hydrolysis of esters is irreversible 

You can't make esters from carboxylic acids and alcohols under basic conditions because the base 
deprotonates the carboxylic acid (p. 000). However, you can reverse that reaction and hydrolyse an 
ester to a carboxylic acid (more accurately, a carboxylate salt) and an alcohol. 






OMe Na0H > H 2° 



100 "C 
5-10 min 




00 Na® 



HCI 




N0 2 

90 - 96% yield 

This time the ester is, of course, not protonated first as it would be in acid, but the unprotonated 

ester is a good enough electrophile because OH~, and not water, is the nucleophile. The tetrahedral 

intermediate can collapse either way, giving back ester, or going forward to acid plus alcohol. 

irreversible deprotonation pulls the equilibrium 
over towards the hydrolysis products 



Ar VOMe 
HO' 



• ■0 






V 



X 

Ar^^O Na® 



Without an acid catalyst, the alcohol cannot react with the carboxylic acid; in fact, the backward 
reaction is doubly impossible because the basic conditions straight away deprotonate the acid to 
make a carboxylate salt (which, incidentally, consumes the base, making at least one equivalent of 
base necessary in the reaction). 



How do we know this is the mechanism? 

Ester hydrolysis is such an important reaction that 
chemists spent a lot of time and effort finding out exactly 
how it worked. If you want to know all the details, read a 
specialist textbook on physical (mechanistic) organic 
chemistry. Many of the experiments that tell us about the 
mechanism involve oxygen-18 labelling. The starting 



material is synthesized using as a starting material a 
compound enriched in the heavy oxygen isotope 18 0. By 
knowing where the heavy oxygen atoms start off, and 
following (by mass spectrometry — Chapter 3) where they 
end up, the mechanism can be established. 



292 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



How do we know this is the mechanism? (continued) 

1 An 18 label in the 'ether' oxygen of the ester starting material ends up in the alcohol product 



x 



H 2 0, OH" 



>OEt 



x 

Me-^^OH 



H 18 0Et 



2 Hydrolysis with 18 0H2 gives 18 0-labelled carboxylic acid, but no 18 0-labelled alcohol 



u 

x - 

mtf OEt 

o (f °" 

INCORRECT 



H 2 18 0, OH" 



HOEt + 



x 



18 0H 



These experiments tell us that a displacement (substitution) has occurred at the carbonyl 
carbon atom, and rule out the alternative displacement at saturated carbon. 

Having worked this out, one further labelling experiment showed that a tetrahedral 
intermediate must be formed: an ester labelled with 18 in its carbonyl oxygen atom passes 
some of its 18 label to the water. We discussed why this shows that a tetrahedral 
intermediate must be formed on p. 000. 



The saturated fatty acid tetradecanoic acid (also known as myristic acid) is manufactured com- 
mercially from coconut oil by base-catalysed hydrolysis. You may be surprised to learn that coconut 
oil contains more saturated fat than butter, lard, or beef dripping: much of it is the trimyristate ester 
of glycerol. Hydrolysis with aqueous sodium hydroxide, followed by reprotonation of the sodium 
carboxylate salt with acid, gives myristic acid. Notice how much longer it takes to hydrolyse this 
branched ester than it did to hydrolyse a methyl ester (p. 000). 



R = 



NaOH, H 2 

* 

100 °c 
p several hours 




Y Y 

o o 

principal component of coconut oil 



X. 



Na(T R 




OH OH 

glycerol 



: C13H27 



HCI 



x 

89-95% 
fatty acid 



Saponification 

The alkaline hydrolysis of esters to give carboxylate salts 
is known as saponification, because it is the process 
used to make soap. Traditionally, beef tallow (the 
tristearate ester of glycerol — stearic acid is octadecanoic 
acid, C17H35CO2H) was hydrolysed with sodium hydroxide 
to give sodium stearate, C 17 H35C0 2 Na, the principal 
component of soap. Finer soaps are made from palm oil 




and contain a higher proportion of sodium palmitate, 
C 15 H 31 C02Na. Hydrolysis with KOH gives potassium 
carboxylates, which are used in liquid soaps. Soaps like 
these owe their detergent properties to the combination of 
polar (carboxylate group) and nonpolar (long alkyl chain) 
properties. 



C0 2 H 



C0 2 H 



octadecanoic acid = stearic acid 



Amides can be hydrolysed under acidic or basic conditions too 

In order to hydrolyse the least reactive of the series of carboxylic acid derivatives we have a choice: we 



Not all carboxylic acid derivatives are equally reactive 



293 



can persuade the amine leaving group to leave by protonating it, or we can use brute force and 
forcibly eject it with concentrated hydroxide solution. 

Amides are very unreactive as electrophiles, but they are also rather more basic than most carboxylic 
acid derivatives: a typical amide has a pi^aH of — 1; most other carbonyl compounds have P-K^s of 
around -7. You might therefore imagine that the protonation of an amide would take place on nitro- 
gen — after all, amine nitrogen atoms are readily protonated. And, indeed, the reason for the basicity of 
amides is the nitrogen atom's delocalized lone pair, making the carbonyl group unusually electron- 
rich. But amides are always protonated on the oxygen atom of the carbonyl group — never the nitrogen, 
because protonation at nitrogen disrupts the delocalized system that makes amides so stable. 

derealization in an un-protonated amide 

n© 




^©/ 

^N 



protonation at 





A, 



:© 



© 



derealization of charge over N and 



protonation at N (does not happen) 
0- H® 



H 




OH 






®y 



no derealization possible 



Protonation of the carbonyl group by acid makes the carbonyl group electrophilic enough for 
attack by water, giving a neutral tetrahedral intermediate. The amine nitrogen atom in the tetrahe- 
dral intermediate is much more basic than the oxygen atoms, so now it gets protonated, and the 
RNH 2 group becomes really quite a good leaving group. And, once it has left, it will immediately be 
protonated again, and therefore become completely nonnucleophilic. The conditions are very vigor- 
ous — 70% sulfuric acid for 3 hours at 100 °C. 

amide hydrolysis in acid: 3 hours at 100 °C with 70% H 2 S0 4 in water gives 70% yield of the acid 



Notice that this means that one 
equivalent of acid is used up in 
this reaction — the acid is not 
solely a catalyst. 



®/H 



^^NHPh Ph^VNHPh Ph- ^ 

u_n — ' 



© 
NH 2 Ph 



H 

H 



protonation of the amine 
prevents reverse reaction 



OH 



© 
PhNH3 



^ x — X 



+ 

PhNH 2 

Vh© 



v OH Ph' 



Hydrolysis of amides in base requires similarly vigorous conditions. Hot solutions of hydroxide 
are sufficiently powerful nucleophiles to attack an amide carbonyl group, though even when the 
tetrahedral intermediate has formed, NH 2 (p^ a H 35) has only a slight chance of leaving when OH~ 
(P-^aH 15) is an alternative. Nonetheless, at high temperatures, amides are slowly hydrolysed by con- 
centrated base. 



amide hydrolysis 
in base 



10% NaOH in H 2 



I 
r^NH, 



(longer for amides of primary 
or secondary amines) 



,^\„© 



1 -J^T - 

R^Y NH 2 IT(NlH 2 

most of the time, hydroxide is lost 
again, giving back starting materials 



baseK 
) f? 



irreversible formation of carboxylate 
anion drives reaction forward 



294 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



Secondary and tertiary amides hydrolyse much more slowly under these conditions. However, 
with a slightly different set of reagents, even tertiary amides can be hydrolysed at room temperature. 



hydrolysis of amides 
using t-BuOK 



.A 



NMe 2 



H 2 (2 equiv.) 
t-BuOK (6 equiv) 

DMSO, 20 °C 

then HCI (to protonate 
carboxylate salt) 



I 

pit^oh 

90% 



Me 2 NH 



85% 



You 've not seen the option of 
as a leaving group before but this 
is what you would get if you want 
0~to leave. Asking 2 " to be a 
leaving group is like asking H0~ to 
bean acid. 



The hydrolysis of some amides in 
aqueous NaOH probably 
proceeds by a similar dianion 
mechanism — see Chapter 13. 



The reason is a change in mechanism. Potassium ferf-butoxide is a strong enough base (piC a H 18) 
to deprotonate the tetrahedral intermediate in the reaction, forming a dianion. Now that the choice 
is between Me 2 N~ and O , the Me 2 N~ has no choice but to leave, giving the carboxylate salt directly 
as the product. 



t-BuO 



™A 



NMe 2 



^ 1 ^\P* 







£)/ 



Ph' 




HO 






NMe 2 PIT l^\NMe 2 ph 

Me 2 N~ has to leave - there's no alternative! 



A, 



e 



x 



-H 2 



Hydrolysing nitriles: how to make the almond extract, mandelic acid 

Closely related to the amides are nitriles. You can view them as primary amides that have lost one 
molecule of water and, indeed, they can be made by dehydrating primary amides. 

They can be hydrolysed just like amides too. Addition of water to the protonated nitrile gives a 
primary amide, and hydrolysis of this amide gives carboxylic acid plus ammonia. 



Don't be put off by the number of steps in this mechanism — look 
carefully, and you will see that most of them are simple proton transfers. 
The only step that isn't a proton transfer is the addition of water. 



.. H 2 0, H 2 S0 4 -. 

PIT ^CN *~ PIT ^C0 2 H 

100 X, 3 h 80% 



, © 



: OH 2 " 

V^P NH xNH N\IH 2 NH 2 

(. ® 



NH 

© 



(ran 
uffi 



reminder: 

cyanohydrins from aldehydes 



HCN 



OH 



RCH0 



CN 



x 

PIT^H 

benzaldehyde 



OH 



You met a way of making nitriles — from HCN (or NaCN + HCI) plus aldehydes — in Chapter 6: 
the hydroxynitrile products are known as cyanohydrins. 

With this in mind, you should be able to suggest a 
way of making mandelic acid, an extract of almonds, 
from benzaldehyde. 

This is how some chemists did it. 



PIT ^C0 2 H 

mandelic acid 



You have just designed your first total 
synthesis of a natural product. We 
return to such things much later in this 
book, in Chapter 31. 



synthesis of 

mandelic acid 

from benzaldehyde 



PhCHO 



NaCN 
* 

u© 



OH 



Ph 



H 2 



OH 



CN 



HCI 



Ph 



C0 2 H 



mandelic acid 
50-52% yield 



Acid chlorides can be made from carboxylic acids using SOCI2 or PCI5 

We have looked at a whole series of interconversions between carboxylic acid derivatives and, after 
this next section, we shall summarize what you should have learned. We said that it is always easy to 
move down the series of acid derivatives we listed early in the chapter and, so far, that is all we have 



Not all carboxylic acid derivatives are equally reactive 



295 



done. But some reactions of carboxylic acids also enable us to move upwards in the series. What 
we need is a reagent that changes the bad leaving group HO - into a good leaving group. Strong 
acid does this by protonating the OH~, allowing it to leave as H2O. In this section we look at two 
more reagents, SOCI2 and PCI5, which react with the OH group of a carboxylic acid and also turn 
it into a good leaving group. Thionyl chloride, SOCI2, reacts with carboxylic acids to make acyl 
chlorides. 



acid chlorides are made 
from carboxylic acids 
with thionyl chloride 



cr xi 



OH 



80 "C, 6 h 



CI 



85% yield 



This volatile liquid with a choking smell is electrophilic at the sulfur atom (as you might expect 
with two chlorine atoms and an oxygen atom attached) and is attacked by carboxylic acids to give an 
unstable, and highly electrophilic, intermediate. 




c|Gr ^ H<; o® 



HO 



CK XI 



A 



XI 






R X"^ XI 

unstable intermediate 



+ HCI 



Protonation of the unstable intermediate (by the HCI just produced) gives an electrophile 
powerful enough to react even with the weak nucleophile CI~ (low p-K" a H> poor nucleophilicity). The 
tetrahedral intermediate that results can collapse to the acyl chloride, sulfur dioxide, and hydrogen 
chloride. This step is irreversible because SO2 and HCI are gases that are lost from the reaction 
mixture. 



You may be shocked to see the 
way we substituted at S=0 
without forming a 'tetrahedral 
intermediate'. Well, this trivalent 
sulfur atom is already tetrahedral 
(it still has one lone pair), and 
substitution can go by a direct 
'S N 2 at sulfur' (Chapter 17). 



A 



0' ^Cl 



unstable intermediate 



HCI 



0/H 



CI 



XI 



CI 






/HClA 
S0 2 

lost from reaction mixture 



Although HCI is involved in this reaction, it cannot be used as the sole reagent for making acid 
chlorides. It is necessary to have a sulfur or phosphorus compound to remove the oxygen. An alter- 
native reagent for converting RCO2H into RCOC1 is phosphorus pentachloride, PCI5. The mecha- 
nism is similar — try writing it out before looking at the scheme below. 





acid chlorides are made 
from carboxylic acids with 
phosphorus pentachloride 



0,N 




PCU 



OoN 




90-96% 
yield 



PCI4 

AC — 

' XH 



X© PCI 4 



^ 



,PCI 4 



H^ © 






CI 



,PCh 




CI 



p 

o""|*N:i 

CI 



0=PCU 



py> J. 



u 



296 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



An alternative method of making acid chlorides: oxalyl chloride plus DMF 



A modification of thethionyl chloride method for 
making acyl chlorides uses oxalyl chloride plus 
catalytic DMF. The oxalyl chloride reacts with the 



DMF in a rather remarkable way to produce a highly 
electrophilic cationic intermediate, plus CO and 



CO2 — as with the SOCI2 reaction, the by-products 
are all gases. 




CI 




© 



CI 



cT Me 



Me 




© 




gaseous by-products 



,Me 



CI 




Me 



reactive intermediate 



A few aspects of this mechanism need comment. 

• The first two steps are simply a nucleophilic 
substitution of CI at the carbonyl group, going via 
the now familiar tetrahedral intermediate 



Nucleophiles can attack the C=N bond (step 3) 
much as they might attack a C=0 bond 
The black arrows in step 4 look very odd, but they 
are the only way we can draw the formation of 
carbon monoxide 



The reactive intermediate is highly electrophilic and 
reacts rapidly with the carboxylic acid, producing 
another intermediate which intercepts CPto give 
the acyl chloride and regenerate DMF. 




-"\£ © 



Me 



Me 



This method is usually used for producing small 
amounts of valuable acyl chlorides — oxalyl chloride 
is much more expensive than thionyl chloride. DMF 




© Me 
N 

I 
Me 



will nonetheless also catalyse acyl chloride 
formation with thionyl chloride, though on a large 
scale its use may be ill advised since one of the 





R' "CI 




A 



Jk /Me 
H N 

I 
Me 



minor by-products from these reactions is a potent 
carcinogen. We hope you enjoyed the eight-step 
mechanism. 



Oxalyl chloride, (COCI) 2 , is the 
'double' acid chloride of oxalic 
acid, orethane-l,2-dioicacid,the 
toxic dicarboxylic acid found in 
rhubarb leaves. 



CI 

v OH Y "CI 

b o 

oxalic acid oxalyl chloride 





These conversions of acids into acid chlorides complete all the methods we need to convert acids 
into any acid derivatives. You can convert acids directly to esters and now to acid chlorides, the most 
reactive of acid derivatives, and can make any other derivative from them. The chart below adds 
reactions to the reactivity order we met earlier. 



acid (acyl) 
chlorides 



anhydrides 



amides 



most reactive 


r^Nji 



R 1 ^ 







R^H 



R^H 



x 



NH 3 



NH 3 



I 



least reactive 



-* 








H20 






S0CI 2 

or 

PCI5 


H20 






or 
(COCI) 2 


1 

H 2 


' 


, 




acid or base 




A ■ 


arboxylic 
cids 


acid only 


© 
R^H, H 






1 


I 




H 2 









strong acid or strong base 



Making ketones from esters: the problem 



297 



All these acid derivatives can, of course, be hydrolysed to the acid itself with water alone or with 
various levels of acid or base catalysis depending on the reactivity of the derivative. To climb the 
reactivity order therefore, the simplest method is to hydrolyse to the acid and convert the acid into 
the acid chloride. You are now at the top of the reactivity order and can go down to whatever level 
you require. 



Making other compounds by substitution reactions of acid 
derivatives 

We've talked at length about the interconversions of acid derivatives, explaining the mechanism of 



attack of nucleophiles such as ROH, H2O, and NH3 on acyl chlorides, acid anhydrides, esters, acids, 
and amines, with or without acid or base present. We shall now go on to talk about substitution reac- 



Five 'oxidation levels' — (1) C0 2 ; (2) 
carboxylic acid; (3) aldehyde and 
ketone; (4) alcohol; and (5) 
tions of acid derivatives that take us out of this closed company of compounds and allow us to make hydrocarbon— were defined in 

compounds containing functional groups at other oxidation levels such as ketones and alcohols. Chapter 2. 



Making ketones from esters: the problem 

Substitution of the OR group of an ester by an R group would give us a ketone. You might therefore 
think that reaction of an ester with an organolithium or Grignard reagent would be a good way of 
making ketones. However, if we try the reaction, something else happens. 






OMe substitution R 1 



r-A. 



A 



MeMgBr 



OMe 



OH 

" R^Me 
Me 



Two molecules of Grignard have been incorporated and we get an alcohol! If we look at the mech- 
anism we can understand why this should be so. First, as you would expect, the nucleophilic 
Grignard reagent attacks the carbonyl group to give a tetrahedral intermediate. The only reasonable 
leaving group is RO~, so it leaves to give us the ketone we set out to make. 



BrMg 



^P 

R'^^OMe 



"*" R'7? s OMe 



Me ^r 



"*" R^^Me 



Now, the next molecule of Grignard reagent has a choice. It can either react with the ester starting 
material, or with the newly formed ketone. Ketones are more electrophilic than esters so the 
Grignard reagent prefers to react with the ketone in the manner you saw in Chapter 9. A stable alkox- 
ide anion is formed, which gives the tertiary alcohol on acid work-up. 



BrMg-^- 



R^^Me 



G 



R 
Me 



.© 



Me 



OH 



R "Me 
Me 



Making alcohols instead of ketones 

In other words, the problem here lies in the fact that the ketone product is more reactive than the 
ester starting material. We shall meet more examples of this general problem later (in Chapter 24, for 
example): in the next section we shall look at ways of overcoming it. Meanwhile, why not see it as a 
useful reaction? This compound, for example, was needed by some chemists in the course of research 
into explosives. 

It is a tertiary alcohol with the hydroxyl group flanked by two identical R (= butyl) groups. The 
chemists who wanted to make the compound knew that an ester would react twice with the same 
organolithium reagent, so they made it from this unsaturated ester (known as methyl methacrylate) 
and butyllithium. 




^^ 



298 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



N<^N>Me 



2 x BuLi 




^^ 



i Tertiary alcohol synthesis 

Tertiary alcohols with two identical R groups can be made from 
ester plus two equivalents of organolithium or Grignard reagent. 



OH 



This reaction works with R=H too if we use lithium aluminium hydride as the source of H . 
L1AIH4 is a powerful reducing agent, and readily attacks the carbonyl group of an ester. Again, col- 
lapse of the tetrahedral intermediate gives a compound, this time an aldehyde, which is more reac- 
tive than the ester starting material, so a second reaction takes place and the ester is converted 
(reduced) into an alcohol, 
reduction of esters by LiAIH 4 



51 



OMe 



"*" R /(^OlVe 



M 1 

©A!H 3 



X 



Q AIH 3 







© 



R / H 
H 



H H 

V 

R^^OH 



This is an extremely important reaction, and one of the best ways of making alcohols from esters. 
Stopping the reaction at the aldehyde stage is more difficult: we shall discuss this in Chapter 24. 

Another bit of shorthand 

Before we go any further, we should introduce to you another little bit of chemical shorthand that 
makes writing many mechanisms easier. 

As you now appreciate, all substitution reactions at a carbonyl group go via a tetrahedral intermediate. 

,D p 

Nu t> 

A convenient way to save writing a step is to show the formation and collapse of the tetrahedral 
intermediate in the same structure, by using a double-headed arrow like this. 

^S? -JL 



Nu o»c 






Nu 




s Nu 



Now, this is a useful shorthand, but it is not a substitute for understanding the true mechanism. 
Certainly, you must never ever write 

xL -X 



Nu 



WRONG 



Nu 



Here's the 'shorthand' at work in the L1AIH4 reduction you have just met. 



J? 

RY^OMe 



.1 



,0 



© 



R / H 
H 



^OH 



Q AlhU 



Q AIH 3 



Making ketones from esters: the solution 



299 



Making ketones from esters: the solution 

We diagnosed the problem with our intended reaction as one of reactivity: the product ketone is 
more reactive than the starting ester. To get round this problem we need to do one of two things: 

1 make the starting material more reactive or 

2 make the product less reactive 

Making the starting materials more reactive 

A more reactive starting material would be an acyl chloride: how about reacting one of these with a 

Grignard reagent? This approach can work: for example, this reaction is successful. 



.MgBr + 




OK ^^ ^-' ^OMe 

81% yield 

Often, better results are obtained by transmetallating (see Chapter 9) the Grignard reagent, or the 

organolithium, with copper salts. Organocopper reagents are too unreactive to add to the product 

ketones, but they react well with the acyl chloride. Consider this reaction, for example: the product 

was needed for a synthesis of the antibiotic septamycin. 

Me Me Me Me 

Me 2 CuLi 

IUIaO 1 1 IUIa 

97% yield 



MeO 




CI 



MeO 




Me 



Notice how this reaction 
illustrates the difference in 
reactivity between an acyl chloride 
functional group and an ester 
functional group. 



OMe 



You met organocopper reagents in 
Chapter 10 where you saw that they did 
conjugate additions to ct.fi-unsaturated 
carbonyl compounds. Other metals, 
such as cadmium or manganese, can 
also be used to make ketones from 
acid chlorides. 



Making the products less reactive 

This alternative solution is often better. With the right starting material, the tetrahedral intermediate can 
become stable enough not to collapse to a ketone during the reaction; it therefore remains completely 
unreactive towards nucleophiles. The ketone is formed only when the reaction is finally quenched with 
acid but the nucleophile is also destroyed by the acid and none is left for further addition. 



acid quench collapses the intermediate and 
simultaneously destroys unreacted organolithium 




R^Li 



.X, 



0® Li® 



.© 



t'-^Qx 



~-A 



choose X carefully and the tetrahedral intermediate is stable 

We can illustrate this concept with a reaction of an unlikely looking electrophile, a lithium car- 
boxylate salt. Towards the beginning of the chapter we said that carboxylic acids were bad elec- 
trophiles and that carboxylate salts were even worse. Well, that is true, but with a sufficiently 
powerful nucleophile (an organolithium) it is just possible to get addition to the carbonyl group of a 
lithium carboxylate. 



Li^efP 



Q Li® 



R^O® Li® 



Me 



0®Li® 



tetrahedral intermediate: 

stable under anhydrous conditions 



We could say that the affinity of lithium for oxygen means that the Li-O bond has considerable 
covalent character, making the CG^Li less of a true anion. Anyway, the product of this addition is a 
dianion of the sort that we met during one of the mechanisms of base-catalysed amide hydrolysis. 
But, in this case, there is no possible leaving group, so there the dianion sits. Only at the end of the 
reaction, when water is added, are the oxygen atoms protonated to give a hydrated ketone, which 
collapses immediately (remember Chapter 6) to give the ketone that we wanted. The water quench 
also destroys any remaining organolithium, so the ketone is safe from further attack. 



300 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



Notice that three equivalents of 
organolithium are needed in this 
reaction: one to deprotonate the 
acid; one to deprotonate the 
hydroxyl group; and one to react 
with the lithium carboxylate. The 
chemists added a further 0.5 for 
good measure! 



tetrahedral 
intermediate 



OLi 

R^OLi 
Me 



3xH 



© 




-H 2 

Me© 



7 Me 



This method has been used to make some ketones that are important starting materials for mak- 
ing cyclic natural products known as macrolides. 



OH 



OH 




C0 2 H 



1. EtLi (3.5 eq) 



2. H® , H 2 




Et 



65% yield 



Another good set of starting materials that leads to noncollapsible tetrahedral intermediates is 
known as the Weinreb amides, after their inventor, S.M. Weinreb. 



Chelation means the 
coordination of more than one 
electron-donating atom in a 
molecule to a single metal atom. 
The word derives from chele, the 
Greek for 'claw'. 

during the reaction: 



JL .OMe 

I 
Me 

a Weinreb amide 
(an N-methoxy-W-methyl amide) 



easily made from 






HN' 

I 
Me 



.OMe 



acyl chloride 



Addition of organolithium or organomagnesium reagents to JV-methoxy-N-methyl amides gives a 
tetrahedral intermediate that is stabilized by chelation of the magnesium atom by the two oxygen 
atoms. This intermediate collapses, to give a ketone, only when acid is added at the end of the reaction. 




OMe 




,0Me 



N 
Me | 
Me 



^MgBr 

V tetrahedral intermediate is 

OMe stabilized by coordination 

1 — ^s/^ N^ °f tne second oxygen atom 
iy| e 1 to the magnesium atom 


KJ 


Me 



on quenching with acid: 



© 




y 



OMe 



N 
Me | 
Me 



© 



-<" 



V 



arkc 
Me |GTH 
Me 



OMe 




summary of reaction 




OMe 



1. MeMgBr 



2. HCI, H 2 




96% yield 



This strategy even works for making aldehydes, if the starting material is dimethylformamide 
(DMF, Me 2 NCHO). 



Me 2 N 



tetrahedral intermediate 



acid added at end 



H 



\£ft stable until... Me N Q of reaction 



©J. 




,© 




tetrahedral 
intermediate 



Me 2}\ \J0H collapse 



-H 



© 




CHO 



And to conclude 



301 



This is an extremely useful way of adding electrophilic CHO groups to organometallic nucleo- 
philes. Here is an example. The first step is an 'ortholithiation' as described in Chapter 9. 
OMe OMe OMe 




NEt, 



s BuLi 




NEt , 



1. Me 2 NCH0 



2. H« 



(iT^ NEt2 

|l .1 75% yield 



A final alternative is to use a nitrile instead of an ester. 
1. PhMgBr 




CN 2.H 3 0© 




Ph 



The intermediate is the anion of an imine (see Chapter 14 for more about imines), which is not 
electrophilic at all — in fact, it's quite nucleophilic, but there are no electrophiles for it to react with 
until the reaction is quenched with acid. It gets protonated, and hydrolyses (we'll discuss this in the 
next chapter) to the ketone. 




MgBr 




Ph 



,© 




e N © h 2 o 

MgBr 



Ph 



To summarize... 

To finish, we should just remind you of what to think about when you consider a nucleophilic sub- 
stitution at a carbonyl group. 



is this carbonyl group 
electrophilic enough? 



^Vx 



Y 

is Y a good enough 
nucleophile? 



tetrahedral 
intermediate 

P 

Y 

which is the better 

leaving group 

XorY? 



is this product more, or less, reactive 
than the starting material? 



A, 



X s 



And to conclude. . . 

In this chapter you have been introduced to some important reactions — you can consider them to be 
a series of facts if you wish, but it is better to see them as the logical outcome of a few simple mecha- 
nistic steps. Relate what you have learned to what you gathered from Chapters 6 and 9, when we first 
started looking at carbonyl groups. All we did in this chapter was to build some subsequent transfor- 
mations on to the simplest organic reaction, addition to a carbonyl group. You should have noticed 
that the reactions of all acid derivatives are related, and are very easily explained by writing out prop- 
er mechanisms, taking into account the presence of acid or base. In the next two chapters we shall see 
more of these acid- and base-catalysed reactions of carbonyl groups. Try to view them as closely 
related to the ones in this chapter — the same principles apply to their mechanisms. 



302 



12 ■ Nucleophilic substitution at the carbonyl (C=0) group 



Problems 



1. Suggest reagents to make the drug 'phenaglycodol' by the route 
shown. HO 

9 ^3^ V 




phenaglycodol 

2. Direct ester formation from alcohols (R OH) and carboxylic 
acids (R CO2H) works in acid solution but does not work at all in 
basic solution. Why not? By contrast, ester formation from alco- 
hols (R OH) and carboxylic acid anhydrides, (R CO^O, or acid 
chlorides, RCOC1, is commonly carried out in the presence of 
amines such as pyridine or Et3N. Why does this work? 

3. Predict the success or failure of these attempted nucleophilic 
substitutions at the carbonyl group. You should use estimated p_K~ a 
or p.JC a H values in your answer and, of course, draw mechanisms. 



x 

Me^^OPh 



n-PrOH 



base 



x 

Me^A)Pr 



x + 

Me^ OPr h 2 N 



A. 




A. 



H 




Me^ N 
H 




? 
HCI 



A, 



4. Suggest mechanisms for these reactions. 



NH 2 



Y 







OH EtO' ^OEt 



^ 



HN^\ 



,COCI 



base 



Y 



A 



5. In making esters of the naturally occurring amino acids (gen- 
eral formula below) it is important to keep them as their 
hydrochloride salts. What would happen to these compounds if 
they were neutralized? 



NH 2 

A 



EtOH 



C0 2 H HC| 



-NH 3 



CI 



C0 2 Et 



6. It is possible to make either the diester or the monoester of 
butanedioic acid (succinic acid) from the cyclic anhydride as 
shown. Why does the one method give the monoester and the 
other the diester? 






OMe 
OMe 



MeOH 




MeO^ 



MeOH 






OH 
OMe 



7. Suggest mechanisms for these reactions, explaining why these 
particular products are formed. 

1 " I I 



Ph 



V CI 



-+- Ph' 



'Ph 



Me 

/ 
-N 



o 



acetone 

1. NaOH 

100 °c 



crystallizes from solution 



H 2 N' 



s C0 2 H 



2. H 



e 



8. Here is a summary of part of the synthesis of Pfizer's heart 
drug Doxazosin (Cordura®). The mechanism for the first step will 
be a problem at the end of Chapter 17. Suggest reagent(s) for the 
conversion of the methyl ester into the acid chloride. In the last 
step, good yields of the amide are achieved if the amine is added as 
its hydrochloride salt in excess. Why is this necessary? 
„C0 2 Me ^^ 0^ 

K 2 C0 3 




OH Br 



Br 



OH 



acetone 




COCI 




C0 2 Me 









/ N 


^s 


\® 


EtOAc 


f 


excess of 


NH 2 


MeOH 


[ 


HCI salt HN. 


J 




H 



Problems 



303 



9. Esters can be made directly from nitriles by acid-catalysed 12. These reactions do not work. Explain the failures and suggest 



reaction with the appropriate alcohol. Suggest a mechanism. 

EtOH 
R — =N —*- R— C0 2 Et 



H 



© 



10. Give mechanisms for these reactions, explaining the selectivi- 
ty (or lack of it!) in each case. 



^\^.° 1. LiAIH 4 ^\^ 1. 
L.0 2. H®H 2 K^O 2. H® 



MeMgl 



H 2 



OCT 



basic solution 

*~ 



acidic solution 




OH 



A 



in each case an alternative method that might be successful. 



e 

.CONH 2 .C0 2 



MeO 



11. This reaction goes in one direction in acidic solution and in 
the other direction in basic solution. Draw mechanisms for the 
reactions and explain why the product depends on the conditions. Me0 2 C, 



^o^ 




OMe HCI 



MeO 




,C0 2 Me 




MeLi 



Equilibria, rates, and mechanisms: 
summary of mechanistic principles 



13 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• Structure of molecules ch4 


• What controls equilibria 


• Kinetics and mechanism ch41 


• Drawing mechanisms ch5 


• Enthalpy and entropy 


• Synthesis in action ch25 


• Nucleophilic attack on carbonyl 


• What controls the rates of reactions 


• How mechanisms are discovered ch41 


groups ch6 & ch9 


• Intermediates and transition states 




• Conjugate addition chlO 


• How catalysts work 




• Acidity and pK a ch8 


• Effects of temperature on reactions 

• Why the solvent matters 





One purpose of this chapter is to help you understand why chemists use such a vast range of dif- 
ferent conditions when performing various organic reactions. If you go into any laboratory, you 
will see many reactions being heated to reflux; however, you will also see just as many being 
performed at -80 °C or even lower. You will see how changing the solvent in a reaction can 
drastically alter the time that a reaction takes or even lead to completely different products. 
Some reactions are over in a few minutes; others are left for hours under reflux. In some reactions 
the amounts of reagents are critical; in others large excesses are used. Why such a diverse range 
of conditions? How can conditions be chosen to favour the reaction we want? To explain all 
this we shall present some very basic thermodynamics but organic chemists do not want to 
get bogged down in algebra and energy profile diagrams will provide all the information we 
need. 



'One could no longer just mix things; 
sophistication in physical chemistry 
was the base from which all 
chemists — including the organic — 
must start.' Christopher Ingold 
(1893-1970) 



How far and how fast? 

We are going to consider which way (forwards or backwards) reactions go and by how much. 
We are going to consider how fast reactions go and what we can do to make them go faster or 
slower. We shall be breaking reaction mechanisms down into steps and working out which step 
is the most important. But first we must consider what we really mean by the 'stability' of mole- 
cules and what determines how much of one substance you get when it is in equilibrium with 
another. 



Stability and energy levels 

So far we have been rather vague about the term stability just saying 
things like 'this compound is more stable than that compound'. What 
we really mean is that one compound has more or less energy than 
another. This comparison is most interesting when two compounds 
can interconvert. For example, rotation about the C-N bond of an 
amide is slow because conjugation (Chapter 7) gives it some double- 
bond character. 

There is rotation, but it can be slow and can be measured by NMR 
spectroscopy. We can expect to find two forms of an amide of the type 
RNH-COR: one with the two R groups trans to one another, and one 
with them cis. 



%* 



180° rotation 



>> 



H 

R groups trans 







I 
H 



R 



it 



H 

R groups cis 



306 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 




Depending on the size of R we should expect one 
form to be more stable than the other and we can 
represent this on an energy profile diagram show- 
ing the relationship between the two molecules in 
energy terms. 



J^^ R 



JO 



0° 90° 180° 

C-N bond rotation 

The two red lines show the 
energies of the molecules and the 
curved black line shows what 
must happen in energy terms as 
the two forms interconvert. 
Energy goes up as the C-N bond 
rotates and reaches a maximum 
at point X when rotation by 90° 
has removed the conjugation. 



rotation 






H 

R groups cis 



90° 




least stable state 
no conjugation 



R groups trans 




K 



JO- 



The relative energies of the two states will 
depend on the nature of R. The situation we have 
shown, with the cis arrangement being much less 
stable than the trans, would apply to large R 
groups. An extreme case would be if the sub- 
stituent on nitrogen were H. Then the two arrange- 
ments would have equal energies. 



x 
k 



0° 



90° 



C-N bond rotation 



180° 



The process is the same but there is now no 
difference between the two structures and, if 
equilibrium is reached, there will be an exactly 
50:50 ratio of the two arrangements. The equi- 
librium constant is K= 1. In other cases, we can 
measure the equilibrium constant by NMR spec- 
troscopy. Another limit is reached if the bond is 
a full double bond as in simple alkenes instead of 
amides. Now the two states do not interconvert. 




c/s-alkene 




frans-alkene 



90° 180° 

C-C bond reaction 



How the equilibrium constant varies with the difference in energy between reactants and products 



307 



We can measure the energies of 
the two molecules by measuring the 
heat of hydrogenation of each isomer 
to give butane — the same product 
from both. The difference between 
the two heats of hydrogenation will 
be the difference in energy of cis- and 
frans-butene. 

In more general terms, amide 
rotation is a simple example of an 
equilibrium reaction. If we replace 
'rotation about the C-N bond' with 
'extent of reaction' we have a picture 
of a typical reaction in which reagents 
and products are in equilibrium. 



J^^ R 




trans 



energy released 
on converting cis 
to trans 



energy released on 

hydrogenating 

trans-butene 



energy released on 

hydrogenating 

c/s-butene 



energy of butane 



JO 




reagents 



products 



extent of reaction 



extent of reaction 



How the equilibrium constant varies with the difference in 
energy between reactants and products 

The equilibrium constant K is related to the energy difference between starting materials and prod- 
ucts by this equation 

AG° = -RT\nK 

where AG° (known as the standard Gibbs energy of the reaction) is the difference in energy between 
the two states (in kj mol ), Tis the temperature (in kelvin not°C), and R is a constant known as the 
gas constant and equal to 8.314 JK mol . 

This equation tells us that we can work out the equilibrium composition (how much of each 
component there is at equilibrium) provided we know the difference in energy between the products 
and reactants. Note that this difference in energy is not the difference in energy between the starting 
mixture and the mixture of products but the difference in energy if one mole of reactants had been 
completely converted to one mole of products. 

Chemical examples to show what equilibria mean 

The equilibrium between isobutyr- 
aldehyde and its hydrate in water 
shows the relationship between AG° 
and K e „. 





308 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



The equilibrium constant may be written to include [H 2 0]; however, since the concentration 
of water effectively remains constant at 55.5 moldm (p. 000), it is often combined into the equi- 
librium constant giving 



The sign of AG" for a reaction tells us 
whether the starting materials or 
products are favoured at equilibrium, 
but it tells us nothing about how long it 
will take before equilibrium is reached. 
The reaction could take hundreds of 
years! This will be dealt with later. 



K, 



[hydrate] eq 



eq 



[aldehyde] eq 

The concentrations of hydrate and aldehyde at equilibrium in water may be determined by 
measuring the UV absorption of known concentrations of aldehyde in water and comparing 
these with the absorptions in a solvent such as cyclohexane where no hydrate formation is possible. 
Such experiments reveal that the equilibrium constant for this reaction in water at 25 °C is 
approximately 0.5 so that there is about twice as much aldehyde as hydrate in the equilibrium mix- 
ture. The corresponding value for AG° is— 8.314 X 298 x ln(0.5) = +1.7 kjmol . In other words, the 
solution of the hydrate in water is 1.7 kjmol higher in energy than the solution of the aldehyde in 
water. 

We could compare this reaction to the addition of an alkyllithium reagent to the same aldehyde. 
You met this reaction in Chapter 9. 




LiO. Me 




MeLi 



The difference in energy between the starting materials, the aldehyde and methyllithium, and the 
products is so great that at equilibrium all we have are the products. In other words, this reaction is 
irreversible. 

The sign of AG° tells us whether products or reactants are favoured at equilibrium 

Consider the equilibrium A^B. The equilibrium constant, K e „, for this reaction is simply given by 
the expression 



Keq = 



[Bl 



teq 



[A], 



eq 



where [A] eq represents the concentration of A at equilibrium. 



If, at equilibrium, there is more B present than A, then _K"will be greater than 1. This means that 
the natural log of K will be positive and hence AG° (given by -RTln K) will be negative. Similarly, if 
A is favoured at equilibrium, K will be less than 1, In K negative, and hence AG° will be positive. If 
equal amounts of A and B are present at equilibrium, _K"will be 1 and, since In 1 = 0, AG° will also be 



• AG" tells us 


about the position 


of equilibrium 




• If AG° 


for a reaction is negative 


, the products will be 


favoured at equilibrium 


• If AG° 


for 


i reaction is positive, 


the reactants will be 


favoured at equilibrium 


• If AG° 
bel 


for 


i reaction is zero, the 


equilibrium constant for the reaction will 



A small change in AG° makes a big difference in K 

The tiny difference in energy between the hydrate and the aldehyde ( 1 .7 kj mol~ ) gave an apprecia- 
ble difference in the equilibrium composition. This is because of the log term in the equation AG° = 
-RTln K: relatively small energy differences have a very large effect on K. Table 13.1 shows the equi- 
librium constants, K e „, that correspond to energy differences, AG°, between and 50 kjmol . These 
are relatively small energy differences — the strength of a typical C-C bond is about 350 kjmol - — 
but the equilibrium constants change by enormous amounts. 



How the equilibrium constant varies with the difference in energy between reactants and products 



309 



In a typical chemical reaction, 'driving 
an equilibrium over to products' might 
mean getting, say, 98% of the products and 
only 2% of starting materials. You can see 
in the table that this requires an equilibri- 
um constant of just over 50 and an energy 
difference of only 10 kJmol~ . This small 
energy difference is quite enough — after 
all, a yield of 98% is rather good! 

Aromatic amines such as aniline 
(PhNH 2 ) are insoluble in water. We saw in 
Chapter 8 that they can be dissolved in 
water by lowering the pH. We are taking 
advantage of the equilibrium between 
neutral amine and its ammonium ion. So 
how far below the p-fC a H of aniline do we 
have to go to get all of the aniline into 
solution? 

If the pH of a solution is adjusted to its 
P_K"aH> by adding different acids there will 
be exactly 50% PhNH 2 and 50% PI1NH3 . 
We need an equilibrium constant of about 
50 to get 98% into the soluble form (PI1NH3) and we need to go only about 2 pK a units below the 
P-KaH of aniline (4.6) to achieve this. All we need is quite a weak acid though in Chapter 8 we used 
HC1 (pK a -7) which certainly did the trick! 

In Chapter 12 (p. 000) we looked at the hydrolysis of esters in basic solution. The decomposition 
of the tetrahedral intermediate could have occurred in either direction as HO~ (pi^aH 15-7) and 
MeO~ (p-K" a H 16) are about the same as leaving groups. In other words K\ and Kj are about the same 
and both equilibria favour the carbonyl compound (ester or carboxylic acid). 



Table 13.1 Variation of K eq 


With AG° 




AG", 


K eq 


%of 


more stable 


kJ mol -1 




state at equilibrium 





1.0 




50 


1 


1.5 




60 


2 


2.2 




69 


3 


3.5 




77 


4 


5.0 




83 


5 


7.5 




88 


10 


57 




98 


15 


430 




99.8 


20 


3 200 




99.97 


50 


580 000 000 


99.9999998 




NH 2 



HX 






NH 3 



,© 



v 

A 



+ HO 







«i 



„0 



OMe 



HOC 

rT^OMe 



ester 



tetrahedral intermediate 






carboxylic acid 







+ MeO 



This reaction would therefore produce a roughly 50:50 mixture of ester and carboxylic acid if this 
were the whole story. But it isn't because the carboxylic acid will be deprotonated in the basic solu- 
tion adding a third equilibrium. 



A 



.© 



HO 



OMe 

ester 






R ^OMe 

tetrahedral intermediate 



A 







MeO 



R ^OH 

carboxylic acid 






© 
R ^0 

carboxylate ion 



MeOH 



Though K\ and K 2 are about the same, K$ is very large (p-fC a of RCO2H is about 5 and p_K" a of 
MeOH is 16 so the difference between the two K a s is about 10 ) and it is this equilibrium that drives 
the reaction over to the right. For the same reason (because K$ is very large), it is impossible to form 
esters in basic solution. This situation can be summarized in an energy diagram showing that the 
energy differences corresponding to JC 1 and K 2 (AG° and AG 2 ) are the same so that AG° between 
RCC^Me + HCF, on the one hand, and RCO2H + MeO - , on the other, is zero. Only the energy dif- 
ference for Kj provides a negative AG° for the whole reaction. 



310 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



k / 








/ 1 


1 ho/ ' 


I \ 




/ AG 1 


V 

R^^OMe 

tetrahedral 
intermediate 


AG° 2 1 y " 




starting materials 


products 


i \ 










° 1 


II 
R^'^^OMe 

* HO® 


x 

R ^OH 

+ MeO 


AG 3 

final 
products 

Ae 

K ^0 








+ MeOH 



extent of reaction 



How to make the equilibrium favour the product you want 

The direct formation of esters 

The formation and hydrolysis of esters was discussed in Chapter 12 where we established that acid 
and ester are in equilibrium and that the equilibrium constant is about one. 



II + MeOH =^i 11 + H 2 



"OH 



"OMe 



If we stew up equal amounts of carboxylic acid, alcohol, ester, and water and throw in a little acid 
to catalyse the reaction (we shall see exactly how this affects the reaction profile later), we find that 
the equilibrium mixture consists of about equal amounts of ester and carboxylic acid. The position 
of the equilibrium favours neither the starting materials or the products. The question now arises: 
how can we manipulate the conditions of the reaction if we actually want to make 100% ester? 

The important point is that, at any one particular temperature, the equilibrium constant is just 
that — constant. This gives us a means of forcing the equilibrium to favour the products (or reactants) 
since the ratio of the two must remain constant. Therefore, if we increase the concentration of the 
reactants (or even that of just one of the reactants), more products must be produced to keep the 
equilibrium constant. One way to make esters in the laboratory is to use a large excess of the alcohol 
and remove water continually from the system as it is formed, for example by distilling it out. This 
means that in the equilibrium mixture there is a tiny quantity of water, lots of the ester, lots of the 
alcohol, and very little of the carboxylic acid; in other words, we have converted the carboxylic acid 
into the ester. We must still use an acid catalyst, but the acid must be anhydrous since we do not want 
any water present — commonly used acids are toluene sulfonic acid (tosic acid, TsOH), concentrated 
sulfuric acid (H2SO4), or gaseous HC1. The acid catalyst does not alter the position of the equilibri- 
um; it simply speeds up the rate of the reaction, allowing equilibrium to be reached more quickly. 

• To make the ester 

Reflux the carboxylic acid with an excess of the alcohol (or the alcohol with an excess of the car- 
boxylic acid) with about 3-5% of a mineral acid (usually HC1 or H2SO4) as a catalyst and distil 
out the water that is formed in the reaction. For example: butanol was heated under reflux with a 



How to make the equilibrium favour the product you want 



311 



fourfold excess of acetic acid and a catalytic amount of concentrated H2SO4 to give butyl acetate 
in a yield of 70%. 



A.. cat. cone. H 2 S0 4 
+ HO^\^\ — ^ J^ 



Me 



^OH 



Me x 



It may also help to distil out the water that is formed in the reaction: diethyl adipate (the diethyl 
ester of hexanedioic acid) can be made in toluene solution using a sixfold excess of ethanol, con- 
centrated H2SO4 as catalyst, distilling out the water using a Dean Stark apparatus. You can tell 
from the yield that the equilibrium is very favourable. 



,.C0 2 H 



cat. cone. H 2 S0 4 



+ EtOH 



„C0 2 Et 



H0 2 C +tiun — Et o 2l 

toluene 96 o /o yie | d 

In these cases the equilibrium is made more favourable by using an excess of reagents and/or 
removing one of the products. The equilibrium constant remains the same. High temperatures 
and acid catalysis are used to speed up arrival at equilibrium which would otherwise take days. 

• To hydrolyse the ester 

Simple: reflux the ester with aqueous acid or alkali. 

The equilibrium between esters and amides 

If you solved Problem 12 at the end of the last chapter, you will already know of one reaction that can 
be driven in either direction by a selection of acidic or basic reaction conditions. The reaction is the 
interconversion of an ester and an 






basic solution 




OH 



acidic solution 



ester 



N 
H 

amide 



X 



amide and one would normally 
expect the reaction to favour the 
amide because of the greater sta- 
bility of amides due to the more 
efficient conjugation of the lone 
pair on nitrogen. 

If we examine the mechanism for the reaction it is clear that ArO~ (p.KaH ~10) is a better leaving 
group than ArNH~ (pJ?aH ~25) and so the equilibria between the two compounds and the tetra- 
hedral intermediate are like this. 




V-A 



NH 2 




N 
H 

tetrahedral intermediate 



V K2 ■ 

/S>H ^- 




OH 



A 



H 

amide 



The two individual equilibria favour the carbonyl compounds over the tetrahedral intermediate 
but K\ < K2 so the overall equilibrium favours the amide. However, two new equilibria must be 
added to these if the variation of pH is considered too. In acid solution the amine will be protonated 
and in base the phenol will be deprotonated. 

ci^=.oa-oo^ca-o0k 

3 2 H H H 

ester tetrahedral intermediate amide 

The energy profile for this equilibrium can be studied from either left or right. It is easiest to 
imagine the tetrahedral intermediate going to the left or to the right depending on the acidity of the 
solution. 



312 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



intermediate 




proton ate d ester 



deprotonated amide 



in acid solution 



in basic solution 



We have shown these last equilibria as reactions because they can be pushed essentially to com- 
pletion by choosing a pH above 10 if we want the amide or below 4 if we want the ester. This is a rel- 
atively unusual situation but there are many other cases where reactions can be driven in either 
direction by choice of conditions. 

Entropy is important in determining equilibrium constants 

The position of equilibrium (that is, the equilibrium constant, which tells us in a chemical reaction 
whether products or reactants are favoured) is determined by the energy difference between the two 
possible states: in the case of the amide RCONH 2 , there is no difference so the equilibrium constant 
is one; in the case of the amide RCONHR with large R groups, the arrangement with R groups trans 
is of lower energy than the state with R groups cis, and so the equilibrium constant is in favour of the 
trans isomer. 

Even when there is a difference in energy between the two states, we still get some of the less stable 
state. This is because of entropy. Why we get the mixture of states is purely down to entropy — there 
is greater disorder in the mixture of states, and it is to maximize the overall entropy that the equilib- 
rium position is reached. 

Energy differences: AG°, AH°, and AS° — energy, enthalpy, and entropy 

Returning to that all important equation: AG° = —RTln K, the sign and magnitude of the energy AG° 
are the only things that matter in deciding whether an equilibrium goes in one direction or another. 
If AG° is negative the equilibrium will favour the products (the reaction goes) and if AG° is large and 
negative the reaction goes to completion. It is enough for AG° to be only about -10 kjmol to get 
complete reaction. The Gibbs energy, AG°, the enthalpy of reaction, AH°, and the entropy of reac- 
tion, AS°, are related via the equation 

AG" = AH - TAS° 



Entropy is important in determining equilibrium constants 313 

The change in enthalpy AH° in a chemical reaction is the heat given out (at constant pressure). Since 
breaking bonds requires energy and making bonds liberates energy, the enthalpy change gives an indi- 
cation of whether the products have more stable bonds than the starting materials or not. Xis the tem- 
perature, in kelvin, at which the reaction is carried out. Entropy, S, is a measure of the disorder in the 
system. A mixture of products and reactants is more disordered than either pure products or pure reac- 
tants alone. AS° represents the entropy difference between the starting materials and the products. 

The equation AG° = AH° - XAS° tells us that how AG° varies with temperature depends mainly on 
the entropy change for the reaction (AS°). We need these terms to explain the temperature depen- 
dence of equilibrium constants and to explain why some reactions may absorb heat (endothermic) 
while others give out heat (exothermic). 

Enthalpy versus entropy — an example 

Entropy dominates equilibrium constants in the difference between inter- and intramolecular reac- 
tions. In Chapter 6 we explained that hemiacetal formation is unfavourable because the C=0 double 
bond is more stable than two C-O single bonds. This is clearly an enthalpy factor depending simply 
on bond strength. That entropy also plays a part can be clearly seen in favourable intramolecular 
hemiacetal formation of hydroxyaldehydes. The total number of carbon atoms in the two systems is 
the same, the bond strengths are the same and yet the equilibria favour the reagents (MeCHO + 
EtOH) in the inter- and the product (the cyclic hemiacetal) in the intramolecular case. 

intermolecular hemiacetal formation intramolecular hemiacetal formation 

OH 





+ ^H ■ > — I 



EtOH + >^ _ — *? HC , 

H 




The difference is one of entropy. In the first case two molecules would give one with an increase in 
order as, in general, lots of things all mixed up have more entropy than a few large things (when you 

drop a bottle of milk, the entropy increases dramatically) . In the second case one molecule gives one There is some discussion f entropy in 
molecule with little gain or loss of order. Both reactions have negative AS° but it is more negative in related reactions in Chapter 6. 
the first case. 

The acidity of chloroacids 

In Chapter 8 we saw how increasing the number of electronegative substituents on a carboxylic acid 
decreased the acid's pK a , that is, increased its acidity. Acid strength is a measure of the equilibrium 
constant for this simple reaction. 






,0 + H® 



R ^OH R' ^0 

carboxylic acid carboxylic acid 

For this equilibrium as for others, the all important equations AG° = -RTlniCand AG° = AH° - 
TAS° apply. When the breakdown of AG° for acid ionization was explored, entropy proved to be 
more important than was expected. Take for example the series CH3COOH, CH2CICOOH, 
CHCl 2 COOH, and CCI3COOH with piC a s 7.74, 2.86, 1.28, and 0.52, respectively. If the increase in 
acidity were simply due to the stabilization of the conjugate base RCO2 by the electronegative 
groups (C-Cl bonds), this would be reflected in the enthalpy difference AH° between the conjugate 
base and the acid. The enthalpy change takes into account the loss of the O-H bond on ionization of 
the acid and also the difference in solvations between the acid and the ions it produces (H bonds 
between RCO2H and water and between RCO2 and water). However the data (see table below) show 
that the difference in equilibrium constant is determined more by entropy than by enthalpy. AH° 
changes by only 6 kj mol - over the whole series while AS° changes by nearly 100 J K~ mol - and the 
more directly comparable TAS changes by over 25 kj mol - . 



314 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



Acid 


P«a 


AH , 


AS , 


-TAS°, 


AG°, 






kJ mol" 1 


JIT^-mor 1 


kJ mol" 1 


kJ mol -1 


CH3COOH 


4.76 


-0.08 


-91.6 


27.3 


27.2 


CH 2 CIC00H 


2.86 


-4.6 


-70.2 


20.9 


16.3 


CHCI 2 C00H 


1.28 


-0.7 


-27 


8.0 


7.3 


CCI3COOH 


0.52 


1.2 


-5.8 


1.7 


2.9 



The entropy change de- 
pends on the difference in 
'order' between the reactants 
and products. Going from one 
species (the undissociated acid) 
to two (the proton and conju- 
gate base) gives an increase in 
entropy. This in turn makes 
AG° more negative and so favours the dissociation. But the solvent structure also changes during the 
reaction. If a species is strongly solvated, it has many solvent molecules tightly associated with it; in 
other words, the solvent surrounding it is more ordered. As a weakly solvated neutral acid ionizes to 
two strongly solvated ions, the neighbouring solvent becomes more ordered and the overall entropy 
decreases. 

As we expect, the p_K" a decreases as more electronegative chlorines are substituted for the hydro- 
gen atoms in acetic acid. However, the enthalpy change for the ionization remains approximately the 
same — the decrease in AG° is predominantly due to the increase in the entropy change for the reac- 
tion. With the increasing numbers of chlorine atoms, the negative charge on the conjugate base is 
more spread out. The less concentrated the charge, the less order is imposed on the neighbouring 
solvent molecules and so AS° becomes less negative. 

Equilibrium constants vary with temperature 

We have said that the equilibrium constant is a constant only so long as the temperature does not 
change. Exactly how the equilibrium constant varies with temperature depends on whether the reac- 
tion is exothermic or endothermic. If the reaction is exothermic (that is, gives out heat) then at high- 
er temperatures the equilibrium constant will be smaller. For an endothermic reaction, as the 
temperature is increased, the equilibrium constant increases. Putting our all important equations 
AG° = -RTlnK and AG° = AH° - TAS° together we see that -RTlnK = AH° - TAS°. If we divide 
throughout by -RTwe have 



Because such trends in pK a are often 
determined by the entropy change of 
the whole system, the order of pK a s 
may change in solvents where there is 
less solvation and be different again in 
the gas phase where there are no 
solvent effects at all. For example, 
whilst the pK a of water is usually 
15.74, in dimethyl sulfoxide (DMSO) it 
is about 29. This is because, in DMSO, 
the hydroxide ion is no longer as 
effectively solvated as it was in water 
and this makes the base much 
stronger. 



InK 



AH" AS" 



RT 



R 



The equilibrium constant JCcan be divided into enthalpy and entropy terms but it is the enthalpy 
term that determines how K varies with temperature. Plotting InK against l/T would give us a 
straight line with slope - AH°/R and intercept AS°. Since T (the temperature in Kelvin) is always pos- 
itive, whether the slope is positive or negative depends on the sign of AH°: if it is positive then, as 
temperature increases, InK (and hence K) increases. In other words, for an endothermic reaction 
(AH positive), as T increases, -K" ([products] /[reactants]) increases which in turn means that more 
products must be formed. 

• Thermodynamics for the organic chemist 

• The free energy change AG° in a reaction is proportional to In K (that is, AG° 
= -RTlnK) 

• AG" and K are made up of enthalpy and entropy terms (that is, AG° = AH° - 

TAS°) 

• The enthalpy change Afl° is the difference in stability (bond strength) of the 
reagents and products 

• The entropy change AS° is the difference between the disorder of the reagents 
and that of the products 

• The enthalpy term alone determines how K varies with temperature 



Making reactions go faster: the real reason reactions are heated 



315 



Le Chatelier's principle 

You may well be familiar with a rule that helps to predict how a system at equilibrium responds to a 
change in external conditions — Le Chatelier's principle. This says that if we disturb a system at equi- 
librium it will respond so as to minimize the effect of the disturbance. An example of a disturbance is 
adding more starting material to a reaction mixture at equilibrium. What happens? More product is 
formed to use up this extra material. This is a consequence of the equilibrium constant being, well... 
constant and hardly needs anybody's principle. 

Another disturbance is heating. If a reaction under equilibrium is heated up, how the equilibrium 
changes depends on whether the reaction is exothermic or endothermic. If is exothermic (that is, 
gives out heat), Le Chatelier's principle would predict that, since heat is consumed in the reverse 
reaction, more of the starting materials will be formed. Again no 'principle' is needed — this change 
occurs because the equilibrium constant is smaller at higher temperatures in an exothermic reaction. 
Le Chatelier didn't know about equilibrium constants or about -RTln K= AH° - TAS° so he needed 
a 'principle'. You know the reasons and they are more important than rules. 

Some reactions are reversible on heating 

Simple dimerization reactions will favour the dimer at low temperatures and the monomer at high tem- 
peratures. Two monomer molecules have more entropy than one molecule of the dimer. An example is 
the dimerization of cyclopentadiene. On standing, 
cyclopentadiene dimerizes and if monomeric material is 
needed the dimer must be heated and the monomer 
used immediately. If you lazily leave the monomer 
overnight and plan to do your reaction tomorrow, you 
will return in the morning to find dimer. 

This idea becomes even more pointed when we look at polymerization. Polyvinyl chloride is the 
familiar plastic PVC and is made by reaction of large numbers of monomeric vinyl chloride mole- 
cules. There is, of course, an enormous decrease in entropy in this reaction and any polymerization 
will not occur above a certain temperature. Some polymers can be depolymerized at high tempera- 
tures and this can be the basis for recycling. 

ow 
temperature 




high 
temperature 



low 
temperature 




This chemistry does not appear until 
Chapter 35 but you do not need to 
know the mechanism of the reaction to 
appreciate the idea. 



cyclopentadiene dimer 



cyclopentadiene 



^S 



CI 



vinyl 
chloride 



high 
temperature 




PVC (polyvinylchloride) 



Polymerization does not appear until 
Chapter 52 but you do not need to 
know the details to appreciate the idea. 



Everything decomposes at a high 
enough temperature eventually 
giving atoms. This is because the 
entropy for lots of particles all 
mixed up is much greater than 
that of fewer larger particles. 



Making reactions go faster: the real reason reactions are heated 

Although in organic laboratories you will see lots of reactions being heated, very rarely will this be to 
alter the equilibrium position. This is because most reactions are not carried out reversibly and so 
the ratio of products to reactants is not an equilibrium ratio. The main reason chemists heat up reac- 
tions is simple — it speeds them up. 

How fast do reactions go? — activation energies 

Using tables of thermodynamic data, it is possible to work out the energy differences for many differ- 
ent reactions at different temperatures. For example, for the combustion of isooctane, AG" (at 298 K) 
= -1000kJmor 1 . 




(I) 



2 (D ^ 



8C0 2 (g) + 9H 2 0(I) AG°=-1000 kJ mol" 



isooctane 



We have seen in Table 13.1 on p. 000 that even a difference of 50 kjmol gives rise to a huge equi 



175 



librium constant: -1000 klmol gives an equilibrium constant of 10 (at 298 K), a number too 



Isooctane (2,2,4-trimethylpentane) is a 
major component of petrol (gasoline). 
Strictly speaking, if we follow the 
standard meaning for 'iso' (p. 000), the 
name isooctane should be reserved for 
the isomer 2-methylheptane. However, 
2,2,4-trimethylpentane is by far the 
most important isomer of octane and 
so, historically, it has ended up with 
this name. 



316 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



vast to contemplate (there are only about 10 atoms in the observable universe). This value of AG° 
(or the corresponding value for the equilibrium constant) suggests that isooctane simply could not 
exist in an atmosphere of oxygen and yet we put it into the fuel tanks of our cars every day — clearly 
something is wrong. 

Since isooctane can exist in an atmosphere of oxygen despite the fact that the equilibrium posi- 
tion really is completely on the side of the combustion products, the only conclusion we can draw 
must be that a mixture of isooctane and oxygen cannot be at equilibrium. A small burst of energy is 
needed to reach equilibrium: in a car engine, the spark plug provides this energy and combustion 
occurs. If no such burst of energy is applied, the petrol would continue to exist for a long time. The 
mixture of petrol and air is said to be kinetically stable but thermodynamically unstable with respect 
to the products of the reaction, CO2 and H2O. If the same small energy burst is applied to the prod- 
ucts, they do not convert back to petrol and oxygen. 

The energy required to overcome the barrier to reaction is called the activation energy and is usu- 
ally given the symbols E a or AG*. An energy level diagram for a reaction such as the combustion of 

E a and AG* are both used for the isooctane is shown below . 

activation energy and are almost 

the same. There are subtle 

differences that do not concern 

us here. 



1 


I 












E a or AG* 


1 




"~X n 


>* 


activation energy for 
forward reaction - 








E a or AG* 


03 


reactants to products 








activation energy for 


0) 


' 


' / 






back reaction - 


01) 

ra 

0) 




reactants 


n \ products to reactants 
AG \ 

\ " 








products 



Svante Arrhenius (1859-1927) was 
one of the founders of physical 
chemistry. He was based at Uppsala in 
Sweden and won the Nobel prize in 
1903 mainly for his theory of the 
dissociation of salts in solution. 



Points to notice: 

• The products are lower in energy than the reactants as the equilibrium position lies in favour of 
the products 

• The activation energy for the forward reaction is less than the activation energy for the back 
reaction 

If a reaction cannot proceed until the reactants have sufficient energy to overcome the activation 
energy barrier, it is clear that, the smaller the barrier, the easier it will be for the reaction to proceed. 
In fact the activation energy is related to how fast the reaction proceeds by another exponential 
equation 

k = Ae RT 

where k is the rate constant for the reaction, R is the gas constant, Tis the temperature (in kelvin), 
and A is a quantity known as the pre-exponential factor. This equation is called the Arrhenius equa- 
tion. Because of the minus sign in the exponential term, the larger the activation energy, E a , the 
slower the reaction but the higher the temperature, the faster the reaction. 

Examples of activation energy barriers 

A very simple reaction is rotation about a bond. In the compounds in the table, different amounts of 
energy are needed to rotate about the bonds highlighted in black. See how this activation energy bar- 
rier affects the actual rate at which the bond rotates. Approximate values for k have been calculated 
from the experimentally determined values for the activation energies. The half-life, t\/2, is just the 
time needed for half of the compound to undergo the reaction. 



Making reactions go faster: the real reason reactions are heated 



317 



NMR spectra of DMF at high and low 
temperature are shown on p. OOO of 
Chapter 7 



We can see how the rate 
constant varies with tempera- 
ture by looking at the 
Arrhenius equation. The pre- 
exponential factor, A, does not 
vary much with temperature, 
but the exponential term is a 
function of temperature. Once 
again, because of the minus 
sign, the greater the tempera- 
ture, the greater the rate con- 
stant. 

This observation is used in 
practice when NMR spectra 
give poor results because of 
slow rotation about bonds. 
Amides of many kinds, partic- 
ularly carbamates, show slow 
rotation about the C-N bond 
at room temperature because 
of the amide derealization. 
These amides have bigger bar- 
riers to rotation than the 70 kJmol~ of the example in the table. The result is a poor spectrum with 
broad signals. In this example, the two sides of the five-membered ring are different in the two rota- 
tional isomers and give different spectra. 



Compound 
H H 

"4: r/ 

c— c«,„ H 

H *H 


Ea, 

kJ mor 1 

12 


Approximate k, 
298 K/s -1 

5xl0 10 


t 1/2 at 298 K 
0.02 ns 


c— c„,„ c| 
cr ti 


45 


8xl0 4 


10u.s 




II 
1 

H 


70 


3 


0.2 s 


Me(l C0 2 Me 
MeT C0 2 Me 


108 


7 x 10~ 7 


11 days 


Ph Ph 

>=< 

H H 


180 


2 x 10 -19 


ca. 10 11 years 3 


a The age of the earth = 4.6 > 


:10 9 years. 







You will see this 'Boc' group used as a 
protecting groupfor amines in Chapter 
24. 




-< »- 




The solution is to run the NMR spectrum at higher temperatures. This speeds up the rotation and 
averages out the two structures. 

A word of warning: heating is not all good for the organic chemist — not only does it speed up the 
reaction we want, it will also probably speed up lots of other reactions that we don't want to occur! 
We shall see how we can get round this, but first we shall take a closer look at what determines how 
fast a reaction takes place. 

Rates of reaction 

Suppose we have the very simple reaction of a single proton reacting with a molecule of water in the 
gas phase 

H + (g) + H 2 0(g) -> H 3 + (g) 

We saw at the beginning of Chapter 8 that this is essentially an irreversible process, that is, AG° is very 
large and negative and therefore the equilibrium constant, K, is large and positive. 

So we know that this reaction goes, but what determines how quickly it can proceed? Since 
the mechanism simply involves one proton colliding with one molecule of water, then clearly the 
rate will depend on how often the two collide. This in turn will depend on the concentrations 
of these species — if there are lots of protons but only a few water molecules, most collisions will be 
between protons. The reaction will proceed fastest when there are lots of protons and lots of water 
molecules. 



This reaction turns two species into 
one, all in the gas phase. The standard 
entropy for the reaction must therefore 
be negative. In order for AG° to be 
negative, the reaction must give out 
heat to the surroundings. In other 
words, this reaction must be highly 
exothermic, as indeed it is. 



318 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



We can express this mathematically by saying that the rate of reaction is proportional to the con- 
centration of protons multiplied by the concentration of water molecules (the square brackets mean 
'concentration of). 

rate of reaction oc [H + ] x [H2O] 

The constant of proportionality, k, is known as the rate constant. 

rate of reaction = kx [H + ] x [H 2 0] 

We are not very interested in reactions in the gas phase, but fortunately reactions in solution fol- 
low more or less the same laws so the reaction of a proton source like HC1 and a water molecule in an 
inert solvent would have the rate expression: rate = kx [HC1] x [H2O]. Expressing the same idea 
graphically requires an energy profile diagram like those we used for equilibria but concentrating 
rather more on AG* than on AG°. 



transition 



l /^~~i 


AG* 


state 


starting materials 






H 2 + HCI 


AG° 




' 


' 




products 
H 3 0® + CI© 



extent of reaction 



Note that the products are lower in energy than the starting materials as before. The energy 
barrier is now marked AG and the highest point on the profile is labelled transition state. Somewhere 
between the starting materials and the products there must come a point where the 
O-H bond is half formed. This is the least stable structure in the whole reaction scheme and would 
correspond to a structure about halfway between starting materials and products, something like this. 



H 2 0: ^H-Qil 

starting materials 



(+) 



H 2 H----C 

transition state 



H 3 0© + 



CI 



,0 



products 



Now notice that the transition state is drawn in square brackets and marked %. Note the 
long dashed bonds not yet completely formed or not yet completely broken and the partial charges 
(+) and (-) meaning something about half a charge (the products have complete charges shown in 
circles). 

• Transition state 

A transition state is a structure that represents an energy maximum on passing 
from reactants to products. It is not a real molecule in that it may have partially 
formed or broken bonds and may have more atoms or groups around the central 
atom than allowed by valence bond rules. It cannot be isolated because it is an 
energy maximum and any change in its structure leads to a more stable 
arrangement. A transition state is often shown by putting it in square brackets 
with a double-dagger superscript. 

This species is unstable — both the starting materials and the products are lower in energy. This 
means that it is not possible to isolate this halfway species; if the reaction proceeds just a little more 
forwards or backwards, the energy of the system is lowered (this is like balancing a small marble on 
top of a football — a small push in any direction and the marble will fall, lowering its potential energy). 



Kinetics 



319 



Kinetics 

The value of the rate constant will be different for different reactions. Consider the reaction of HC1 
and a water molecule discussed in the last section. Even with the same concentrations, the almost 
identical reaction where hydrogen is replaced by deuterium will proceed at a different rate (Chapter 
19). To understand this we need to think again about what needs to happen for a reaction to occur. It 
is not enough for the two species to simply collide. We know that for this reaction to work the proton 
must come into contact with the oxygen atom in the water molecule, not the hydrogen atoms, that is, 
there is some sort of steric requirement. We have also seen that most reactions need to overcome an 
energy barrier. In other words, it is not enough for the two species just to collide for a reaction to 
proceed, they must collide in the right way and with enough force. 

You can see now how the overall rate equation for our example reaction 

rate of reaction = kx [HCI] x [H 2 0] 

contains all the points needed to work out how fast the reaction will proceed. The most important 
point concerns the concentrations of the reacting species — which are expressed directly in the rate 
equation. Other considerations, such as how large the species are or whether or not they collide in 
the right way with the right energy, are contained in the rate constant, k. Notice once again that not 
only is k different for different reactions (for all of the above reasons), but that it also varies with tem- 
perature. It is essential when quoting a rate constant that the temperature is also quoted. That part of 
chemistry that deals with reaction rates rather than equilibria is known as kinetics. 

Activation barriers 



In the same way that we define AG* to be the 
difference in energy between the starting materials 
and the transition state (that is, activation energy), 
we can define the entropy of activation , AS*, and the 
enthalpy of activation, AH*, as being the entropy and 
enthalpy differences between the starting materials 
and transition sate. These quantities are directly 
analogous to the entropy and enthalpy of the 
reaction but instead referto the difference between 
starting material and transition state rather than 
starting material and products. 

In a similar manner, we could also define an 
equilibrium constant between the reactants and the 
transition state 



K* 



[AB] 



[A][B] 

Our all-important thermodynamic equations apply 
equally well to these activation functions so that we 
may write 

AG* = -RTInK* and AG* = AH* - TASK 



It is possible to relate these functions with the rate 
constant for the reaction, k, by using a model known 
as transition state theory. We will not go into any 
details here, but the net result is that 



AB* transition state 



h 



-K' : 



where kg and h are universal constants known as 

Botlzmann's constant and Planck's constant, 

respectively 

t ~— ' 
By substituting in the equation K + = e Kr the 

rearranged form of AG* = -R7"lnK*) we arrive at an 

equation, known as the Eyring equation, which 

relates how fast a reaction goes (k) to the activation 

energy (AG*) 

k T -— 
k = ^-e "t 
h 

This can be rearranged and the numerical values of 
the constants inserted to give an alternative form 

AG* (in J moP 1 ) = 8.314 x Tx [23.76 + \n{T/k)] 




A+ B 
starting materials 



products 



Kinetics gives us an insight into the mechanism of a reaction 

Now for some of the reactions you have seen in the last few chapters. Starting with carbonyl substitu- 
tion reactions, the first example is the conversion of acid chlorides into esters. The simplest mecha- 
nism to understand is that involved when the anion of an alcohol (a metal alkoxide RO~) reacts with 
an acid chloride. The kinetics are bimolecular: rate = A:[MeCOCl] [RO~] . The mechanism is the simple 
addition elimination process with a tetrahedral intermediate. 

The formation of the tetrahedral intermediate by the combina- 
tion of the two reagents is the rate-determining step and so the 
highest transition state will be the one leading from the starting 
materials to that intermediate. © 



o — ' 



rate- 
determining 

step V^. _ 






V 



+ CI 







RO 



320 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



transition 
1 \ ~" " state 



ntermediate 




starting materials 



+ R0 



^0R + 



CI 







extent of reaction 



We shall return to this important mechanism in a moment after a brief mention of first-order 
kinetics. The reaction between the acid chloride and the neutral alcohol to give an ester may not have 
the bimolecular rate expression expected for this mechanism: rate = k[R COC1] [R OH]. 



^\a 



acid chloride 



R 2 OH 



u 

x 

1 ^^0R 2 



ester 



expected C*"9 

mechanism ^Ml 

note shorthand 



Ri^Scf 



R 2 0H 



Some such reactions have a simpler rate expression: rate = k[R COC1] in which the alcohol does not 
appear at all. Evidently, no collision between the acid chloride and the alcohol is required for this reac- 
tion to go. What actually happens is that the acid chloride decomposes by itself to give a reactive cation 
(a cation you have already seen in mass spectrometry) with the loss of the good leaving group Cl~. 

© 






rate-determining 
step 



a 



fast 



A- 



fast 



R 2 0H 



OR^ 

I 
H 



1 ^^0R 2 



There are three steps in this reaction scheme though the last is a trivial deprotonation. Evidently, 
the energy barrier is climbed in the first step, which involves the acid chloride alone. The cation is an 
intermediate with a real existence and reacts later with the alcohol in a step that does not affect the 
rate of the reaction. The easiest way to picture this detail is in an energy profile diagram (top right). 

Points to notice: 

• The products are again lower in energy than the starting materials 

• There are three transition states in this reaction 

• Only the highest-energy transition state matters in the reaction rate (here the first) 

• The step leading to the highest transition state is called the rate-determining step 

• The two intermediates are local minima in the reaction profile 

• The highest- energy transition state is associated with the formation of the highest-energy inter- 
mediate 



Kinetics 



321 





^\* 




transition 
state 












© 


AG+ 


^N 


- intermediate 








F 


1 















R 1 ^ ^OR 2 


--V-- intermediate 


starting r 


naterials 






I 
R^^CI 


AG° 

r 




1 
H 














products 


x 

R^^OR 2 



extent of reaction 

• Intermediates and transition states 

A transition state represents an energy maximum — any small displacement leads 
to a more stable product. An intermediate, on the other hand, is a molecule or ion 
that represents a localized energy minimum — an energy barrier must be overcome 
before the intermediate forms something more stable. As you have seen in Chapter 
3, and will see again in Chapter 22, because of this energy barrier, it is even possible 
to isolate these reactive intermediates (RCO + ) and study their spectra. 

Because the rate- determining step involves just one molecule, the rate equation shows rate = 
k[R COC1], and the reaction is called a first-order reaction as the rate is proportional to just one 
concentration. A first-order reaction involves the unimolecular decomposition of something in the 
rate-determining step. 

Second-order reactions 

The unimolecular mechanism is unusual for carbonyl substitution reactions. Those in the last chap- 
ter as well as the carbonyl addition reactions in Chapter 6 all had nucleophilic addition to the car- 
bonyl group as the rate- determining step. An example would be the formation of an ester from an 
anhydride instead of from an acid chloride. 
P P ROH 

*► II + MeC0 2 H 

s 0' ^ ^N)R 

The leaving group (MeCO^) is not now good enough (pi>C a H about 5 instead of -7 for Cl~) to leave 
of its own accord so the normal second-order mechanism applies. The kinetics are bimolecular: rate = 
A:[(MeCO)20] [ROH] and the rate-determining step is the formation of the tetrahedral intermediate. 






^A 



rate- 
determining 
step 



ROH 




© 
OH 



*~ ^N)R ^^ ^N>R 



322 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



All the acid derivatives (acid chlorides, anhydrides, esters, and amides) combine with a variety of 
nucleophiles in very similar bimolecular mechanisms. 



X = CI 

OCOR 

OR 

orNH 2 




RO 



rate- g, 

determining [ *) 
step V_^. 



2>Q 



' ^OR 



OR 



This is the simplest and the most typical bimolecular mechanism with one intermediate, and the 
energy profile diagrams are correspondingly easier to understand. The reactions with acid chlorides 
(discussed a few pages back) and anhydrides are straightforward and go in good yield. 




ntermediate 



products 


1 Cl 








intermediate 



products 



© 
MeC0 2 



extent of reaction 



extent of reaction 



The energy levels of the starting materials, the transition state, and the intermediate are all 
lower in the anhydride reaction than in the acid chloride reaction. So which goes faster? We know 
the answer — acid chlorides are more reactive than anhydrides towards nucleophiles. The reason is 
that the stability of the starting materials is determined by the interaction between the carbonyl 
group and the substituent attached directly to it. This is a big effect as we know from infrared 
spectroscopy. 

The two intermediates also have different energies depending mainly on the stability of the 
oxyanion. This too will be affected by the substituents, Cl and OAc, but they are separated from the 
oxyanion by the tetrahedral carbon atom and there is no conjugation. Substituent effects on the 
oxyanion are smaller than they are on the starting materials so the two intermediates are similar in 
energy. Substituent effects on the transition state will be somewhere between the two but the transi- 
tion state is nearer to the intermediate than to the starting material so substituent effects will be like 
those on the intermediate. The two transition states also have similar energies. The net result is that 
AG is bigger for the anhydride mainly because the energy of the starting materials is lower. This also 
explains why AG° is smaller. 



The ester exchange reaction 

When we move on to esters 
reacting with alkoxides the 
chart is a good deal more 
symmetrical. This is the 
reaction. 



3- 







x 

^^OR 1 



R-=0 



© 



R x 



OR 1 



Catalysis in carbonyl substitution reactions 



323 



The nucleophile and the leaving group are both alkoxides, the only difference being R and R . If 
R and R were the same, the energy profile diagram would be totally symmetrical and small differ- 
ences between R and R are not going to affect the symmetry much. 

transition 
state 



So 




intermediate 



starting materials 



RIO® 







R"0 



extent of reaction 

Points to notice: 

• The transition states for the two steps are equal in energy 

• AG is the same for the forward and the back reaction 

• AG° is zero 

• IfR =R , the intermediate has an exactly 50% chance of going forward or backward 

In fact, we now have an equilibrium reaction. If R and R are different then the reaction is called 
ester exchange or transesterification and we should drive it in the direction we want by using a large 
excess of one of the two alcohols. If we carried out the o 

reaction on one ester using an equivalent of the other 
alkoxide in that alcohol as solvent, the other ester 
would be formed in good yield. 



^^OR 2 



R x O 







R x OH 






Catalysis in carbonyl substitution reactions 

We don't need the equivalent of alkoxide in ester exchange because alkoxide is regenerated in the 
second step. We need only catalytic quantities (say, 1-2% of the ester) because the role of the alkox- 
ide is catalytic. It speeds up the reaction because it is a better nucleophile than the alcohol itself and it 
is regenerated in the reaction. 



SI 

R x O ~* 



R^H 







o 



OR 1 



Making a solution more basic speeds up reactions in which alcohols act as nucleophiles because it 
increases the concentration of the alkoxide ion, which is more nucleophilic than the alcohol itself. 
The same thing happens in hydrolysis reactions. The hydrolysis of esters is fast in either acidic or 
basic solutions. In basic solution, hydroxide is a better nucleophile than water. 



324 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



You will also see rate constants 
labelled in other ways — this is a 
matter for choice. A common 
method is to use k± for 
unimolecular and k^ for 
bimolecular rate constants. 



^ 



H 2 



r% o o 



e 



OH 



The mechanism is like that for ester exchange but hydroxide is used up in deprotonating the car- 
boxylic acid produced so a whole equivalent of NaOH is needed. In acidic solution, protonation of 
the carbonyl oxygen atom makes the ester more electrophilic and attack by the weak nucleophile 
(water) is made faster but the acid catalyst is regenerated. In both these reactions nucleophilic attack 
is the rate-determining step. 



© 



^^OR H,0 




^OH 



So, the higher the concentration of protons, the faster the hydrolysis goes and, the higher the con- 
centration of hydroxide ion, the faster the reaction goes. If we plot the (log of the) rate of the reaction 
against the pH of the solution we shall get two straight lines increasing at high and low pH and each 
with a slope of one. The lines intersect near neutrality when there are neither protons nor hydroxide 
ions. This is simple acid and base catalysis. 



log rate 



variation of rate of ester hydrolysis with pH 



u u 

X — -X 

^\ R H 2 ^\ H 



ROH 




These are bimolecular reactions with bimolecular kinetics and the rate expression in each case 
includes the concentration of the catalyst. We can label the rate constants k a and k\, with a suffix 'a' 
for acid and 'b' for base to show more clearly what we mean. 

rate of ester hydrolysis in acid solution (pH < 7) = /c a [MeC0 2 R][H30 + ] 

rate of ester hydrolysis in basic solution (pH > 7) = /^[MeCC^RHHCT] 

Catalysis by weak bases 

In Chapter 12 pyridine was often used as a catalyst in carbonyl substitution reactions. It can act in 
two ways. In making esters from acid chlorides or anhydrides pyridine can act as a nucleophile as 
well as a convenient solvent. It is a better nucleophile than the alcohol and this nucleophilic catalysis 
is discussed in Chapter 12. But nonnucleophilic bases also catalyse these reactions. For example, 
acetate ion catalyses ester formation from acetic anhydride and alcohols. 






ROH 



R0^\ 







Could this be nucleophilic catalysis too? Acetate can certainly attack acetic anhydride, but the 
products are the same as the starting materials. This irrelevant nucleophilic behaviour of acetate ion 
cannot catalyse ester formation. 



The hydrolysis of amides can have termolecular kinetics 



325 



A 5^ A — M&K — A, * A A 



starting materials same as "products" 



tetrahedral intermediate 



"products" same as starting materials 



Can acetate be acting as a base? With a pi^H of about 5 it certainly cannot remove the proton 
from the alcohol (piC a H about 15) before the reaction starts. What it can do is to remove the proton 
from the alcohol as the reaction occurs. 





rate-determining 
step 



-RO^A 



o 



tetrahedral intermediate 



This type of catalysis, which is available to any base, not only strong bases, is called general base 
catalysis and will be discussed more in Chapters 41 and 50. It does not speed the reaction up very 
much but it does lower the energy of the transition state leading to the tetrahedral intermediate since 
that intermediate is first formed as a neutral compound instead of a dipolar species. Here is the 
mechanism for the uncatalysed reaction. 



c*a o 



rate- 
determining 
step 



ROH 




C OH 



© 
OH 



■*- ^N>R 



catalyst 
regenerated 



^N)R 



The disadvantage of general base catalysis is that the first, rate-determining, step is termolecular. 
It is inherently unlikely that three molecules will collide with each other simultaneously and in the 
next section we shall reject such an explanation for amide hydrolysis. In this case, however, if ROH is 
the solvent, it will always be present in any collision so a termolecular step is just about acceptable. 



The hydrolysis of amides can have termolecular kinetics 

When we come to reactions of amides we are at the bottom of the scale of reactivity. Because of the 
efficient derealization of the nitrogen lone pair into the carbonyl group, nucleophilic attack on the 
carbonyl group is very difficult. In addition the leaving group (NHJ, P-KaH about 35) is very bad 
indeed. 



J 



slow step 



-Ai : 



very 
slow step 



A 0H * *« e = A e . 



NH 3 



This reaction was first discussed in 
Chapter 12. 



You might indeed have guessed from our previous example, the hydrolysis of esters, where the 
transition states for formation and breakdown of the tetrahedral intermediate had about the same 
energies, that in the hydrolysis of amide the second step becomes rate-determining. This offers the 
opportunity for further base catalysis. If a second hydroxide ion removes the proton from the tetrahe- 
dral intermediate, the loss of NH2 is made easier and the product is the more stable carboxylate ion. 



SL 



HO 



A"' 



HO 




r% 







rate- 
determining 
step 



A, 



© + H 2 N 



© 



Of course the very basic leaving 
group (NH2, pK aH about 35) 
instantaneously reacts with water 
(pK a H about 15) in a fast proton 
transfer to give NH 3 and HO~. 



326 13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 

Notice that in the first mechanism the hydroxide is consumed as the product eventually 
emerges as an anion. In the second mechanism, one hydroxide is consumed but the second is catalytic 
as the NH2 reacts with water to give ammonia and hydroxide ion. The rate expression for the hydroly- 
sis of amides includes a termolecular term and we shall label the rate constant £3 to emphasize this, 
rate = /< 3 [MeC0NH 2 ][H0-] 2 

Where do the termolecular kinetics come from? It is, of course, extremely unlikely that three species 
will collide simultaneously, particularly as two of them are mutually repelling anions. The rate-deter- 
mining step is actually unimolecular — the spontaneous breakdown of a dianion. But the concentration 
of the dianion is in the rate expression too and that depends on the reactions before the rate-determin- 
ing step. With a late rate-determining step, the previous steps are in equilibrium and 
so we can put in some rate and equilibrium constants for each step and label the intermediates like this. 



i^ 



^Vnh 2 . - .^ ^^ Ma -x'kv 

) ^^NH 2 ^PKiH 2 rate ° 

H <r r~^»<\ & dete ™ g 



© + H 2 N® 



step 
amide HO" monoanion dianion 



The rate of the reaction is the rate of the rate-determining step 

rate = /([dianion] 
We don't know the concentration of the dianion but we do know that it's in equilibrium with the 
monoanion so we can write 
[dianion] 
2 [monoanion][HO~] 
and so [dianion] = K2[monanion][H0~] 
In the same way we don't want the unknown [monoanion] in our rate expression and we can get 
rid of it using the first equilibrium 
[monoanion] 
1 " [amide][HCT] 

and so [monoanion] = K^amide^HCT] 
Substituting these values in the simple rate equation we discover that rate = ^[dianion] becomes 

rate = /<K 1 K 2 [amide][H0-] 2 
The termolecular kinetics result from two equilibria starting with the amide and involving two 
hydroxide ions followed by a unimolecular rate-determining step, and the 'termolecular rate constant' k?, 
is actually a product of the two equilibrium constants and a unimolecular rate constant A3 = kx K^x Kj. 
We have now seen examples of unimolecular and bimolecular reactions and also how termolecu- 
lar kinetics can arise from unimolecular and bimolecular reactions. 

Just because a proposed mechanism gives a rate equation that fits the experimental data, it does 
not necessarily mean that it is the right mechanism; all it means is that it is consistent with the experi- 
mental facts so far but there maybe other mechanisms that also fit. It is then up to the experimenter to 
design cunning experiments to try to rule out other possibilities. 

Mechanisms are given throughout this book — eventually you will learn to predict what the mech- 
anism for a given type of reaction is, but this is because earlier experimentalists have worked out the 
mechanisms by a study of kinetics and other methods (see Chapter 41 for more details on how 
mechanisms are elucidated). In Chapter 17 you will meet another pair of mechanisms — one first- 
order and one second-order — following the same pattern as these. 

The cis- trans isomerization of alkenes 

The fact that a reaction is favourable (that is, AG° is negative) does not mean that the reaction 
will go at any appreciable rate: the rate is determined by the activation energy barrier that must be 



The cis-trans isomerization of alkenes 



327 



overcome. Returning to the example of the cis-trans isomerism of butene, the energy difference 
between two forms is just 2 kj mol ; the activation energy barrier is much bigger: 260 kj mol . The 
difference in energy determines the equilibrium position (2 kJmol~ corresponding to an equilibri- 
um constant of about 2.2, or a ratio of 30:70, cis:trans; see table on p. 000), whilst the activation ener- 
gy determines how fast the reaction occurs (260 kj mol~ means that the reaction does not happen at 
all at room temperature). A calculation predicts that the half- life for the reaction would be approxi- 
mately 10 years at room temperature, a time interval much greater than the age of the universe. At 
500 °C, however, the half- life is a more reasonable 4 hours which just goes to show the power of 
exponentials! Unfortunately, when most alkenes are heated to these sorts of temperatures, other 
unwanted reactions occur. 

In order to interconvert the cis and trans isomers we must use a different strategy. One method is 
to shine light on the molecule. If UV light is used it is of the right wavelength to be absorbed by the 
C=C 7t bond exciting one of the 7t electrons into the antibonding 7t* orbital. There is now no 7t bond 
and the molecule can rotate freely. 



r/^ F 



trans (£-) alkene 



hv 



The quantum symbols hv are 
conventionally used for light in a 
reaction and the excited state 
diagram is the best we can do for 
a molecule with one electron in 
the k orbital and one in the k* 
orbital. 



alkene excited state 



cis (Z-) alkene 



Another approach to alkene isomerization would be to use a catalyst. Base catalysis is of no use as 
there are no acidic protons in the alkene. Acid catalysis can work (Chapter 19) if a carbocation is 
formed by protonation of the alkene. 

How to catalyse the isomerization of alkenes 

The rate at which a reaction occurs depends on its activation energy — quite simply, if we can 
decrease this, then the reaction rate will speed up. There are two ways by which the activation energy 
may be decreased: one way is to raise the energy of the starting materials; the other is to lower the 
energy of the transition state. In the cis/trans isomerization of alkenes, the transition state will be 
halfway through the twisting operation — it has p orbitals on each carbon at right angles to each 
other. It is the most unstable point on the reaction pathway. 



transition state of 
uncatalysed reactior 

transition state of 
catalysed reaction 




trans 



extent of reaction 
Lowering the energy of the transition state means stabilizing it in some way or other. For example, 
if there is a separation of charge in the transition state, then a more polar solvent that can solvate this 
will help to lower the energy of the transition state. Catalysts generally work by stabilizing the transi- 
tion states or intermediates in a reaction. We shall return to this point when we have introduced 
kinetic and thermodynamic products. 



328 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



Kinetic versus thermodynamic products 

In Chapter 10 we discussed conjugate addition to unsaturated carbonyl compounds in contrast to 
direct addition to the carbonyl group. A classic illustration is the addition of HCN to butenone. Two 
products can be formed. 




HCN 



CH 3 

direct addition 
kinetic product 



fast 




HCN 



slow 



CN CH 3 

conjugate addition 
thermodynamic product 



The 'direct' addition to the left means that cyanide ion must attack the carbonyl group directly 
while the 'conjugate' addition to the right means that it must attack the less electrophilic alkene. The 
second is a slower reaction but gives the more stable product. Both reactions have an alkoxide anion 
as an intermediate. 




« c°? 







© 



CN 



CH a 



CN 



CH 3 



CN 



The energy profile diagram for these two reactions is quite complicated. It has the starting mater- 
ial in the middle, as in the mechanism above, and so extent of reaction increases both to the right for 
thermodynamic control and to the left for kinetic control. 

kinetic control thermodynamic control 



intermediate 




thermodynamic 
products 





CN CH 3 



extent of reaction 

Points to notice: 

• The thermodynamic product has a lower energy than the kinetic product 

• The highest transition state to the right is higher than the highest to the left 

• Initially the reaction will go to the left 



Kinetics versus thermodynamic products 



329 



• If there is enough energy for the kinetic product to get back to the starting materials, there will be 
enough energy for some thermodynamic product to be formed 

• The energy needed for the thermodynamic product to get back to starting materials is very great 

• The kinetic product is formed reversibly; the thermodynamic product irreversibly 

• At low temperatures direct addition is favoured, but conjugate addition is favoured at high tem- 
peratures 

Kinetic versus thermodynamic control in the isomerization of alkenes 

Our catalyst for the isomerization of alkenes is going to be HC1 absorbed on to solid alumina (alu- 
minium oxide, AI2O3) and the isomerization is to occur during a reaction, the addition of HC1 to an 
alkyne, in which the alkenes are formed as products. In this reaction the oxalyl chloride is first mixed 
with dried alumina. The acid chloride reacts with residual water on the surface (it is impossible to 
remove all water from alumina) to generate HC1, which remains on the surface. 



oxalyl chloride 



Al,0 3 



+ H 2 



2HCI 
on Al 2 3 



+ CO, 



+ CO 



The treated alumina with HC1 still attached is added to a solution of an alkyne (1-phenylpropyne) 
and an addition reaction occurs to produce two geometrical isomers of an alkene. One results from 
cis addition of HC1 to the triple bond, and one from trans addition. 

HCI from (C0CI) 2 Cl \ / H 

*- X* + 



Ph 



-CH 3 



Al 2 3 



Ph' 



CH 3 



Prf H 



l-phenylpropyne 



E-alkene 
from cis addition of HCI 



Z-alkene 
from trans addition of HCI 



The two alkenes are labelled E and Z . After about 2 hours the main product is the Z-alkene. 
However, this is not the case in the early stages of the reaction. The graph below shows how the pro- 
portions of the starting material and the two products change with time. 

Points to note: 



When the alkyne concentration drops almost 
to zero (10 minutes), the only alkene that has 
been formed is the E-alkene 

As time increases, the amount of E-alkene 
decreases as the amount of the Z-alkene 



Z-alkene 



increases 



of E- and 




• Eventually, the proportions 
Z-alkenes do not change 

Since it is the Z-alkene that dominates at equilibrium, this must be lower in energy than the E- 
alkene. Since we know the ratio of the products at equilibrium, we can work out the difference in 
energy between the two isomers 

ratio of EZalkenes at equilibrium = 1:35 
[Z] 



K eq - [E] 



35 



AG" = -RT\n K= -8.314 x 298 x ln(35) = -8.8 kJ mol" 1 

that is, the Z-alkene is 8.8 kJ mol~ lower in energy than the E-alkene. 

Since the £- alkene is the quickest to form under these conditions, cis addition of HCI must have a 
smaller activation energy barrier than trans addition. This suggests that reaction occurs on the sur- 
face of the alumina with both the H and the CI added to the triple bond simultaneously from the 
same side rather like a's-hydrogenation of triple bonds on a palladium catalyst (p. 000). 



You might normally expect an E- 
alkene to be more stable than a Z- 
alkene — it just so happens here 
that CI has a higher priority than 
Ph and the Z-alkene has the two 
largest groups (Ph and Me) trans. 
(See p. 000 for rules of 
nomenclature.) 



330 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



Ph- 



-CH-, 



CI H 

alumina surface 



Ph- 



-CH 3 



CI- - -H 

alumina surface 



Ph 
CI 



>- 






CH, 



alumina surface 



There must then be some mechanism by which the quickly formed £- alkene is converted into the 
more stable Z-alkene, presumably through another intermediate that is more stable than the transi- 
tion state for alkene interconversion. This information is summarized on a reaction profile diagram. 



transition state 

for HCI addition 

on alumina 




Z-alkene 



extent of reaction 

Initially, the alkyne is converted into the fj-alkene. The activation energy for this step is labelled 
AGf. The .E-alkene then converts to the Z isomer via an intermediate. The activation energy for this 
step is AG2. Overall, the reaction is the addition of HCI to the alkyne to give the Z-alkene — we could 
look on the E isomer as just another intermediate. The only difference between the .E-alkene and the 
intermediate in the isomerism reaction is the size of the activation energies; it is much easier to iso- 
late the ii-alkene because the activation energies to be overcome (AG2 and AG|) are both much larg- 
er than those of the intermediate (AG5 and AG3). The activation energy to be overcome to form the 
£-alkene (AGj) is less than that to be overcome to form the Z-alkene (AG|). 

So what is this intermediate in the isomerization reaction? It is a cation from protonation of the 
alkene by more HCI. The cation is stabilized by derealization into the benzene ring and can rotate as 
it has no double-bond character. 




E-alkene 



stabilized (conjugated) carbocation 



Z-alkene 



1 Kinetic and thermodynamic products 

The £-alkene is formed faster and is known as the kinetic product; the Z-alkene is 
more stable and is known as the thermodynamic product. 



Low temperatures prevent unwanted reactions from occurring 



331 



If we wanted to isolate the kinetic product, the E-alkene, we would carry out the reaction at low 
temperature and not leave it long enough for equilibration. If, on the other hand, we want the ther- 
modynamic product, the Z-alkene, we would leave the reaction for longer at higher temperatures to 
make sure that the larger energy barrier yielding the most stable product can be overcome. 



Low temperatures prevent unwanted reactions from 
occurring 

So far in this chapter we have seen why chemists heat up reaction mixtures (usually because the reac- 
tion goes faster) but in the introduction we also said that, in any organic laboratory, an equal num- 
ber of reactions are carried out at low temperatures. Why might a chemist want to slow a reaction 
down? Actually, we already hinted at the answer to this question when we said that it is possible to 
isolate reactive carbocations. It is possible to isolate these reactive intermediates but only at low tem- 
peratures. If the temperature is too high then the intermediate will have sufficient energy to over- 
come the energy barrier leading to the more stable products. 

In our discussion of the reactions of acid chlorides, we deduced that a unimolecular reaction to 
give a cation must be happening. This cation cannot be detected under these conditions as it reacts 
too quickly with nucleophiles. If we remove reactive 
nucleophiles from solution, the cation is still too 
unstable to be isolated at room temperature. But if we 
go down to -120 °C we can keep the cation alive long 
enough to run its NMR spectrum. 

Lowering the temperature lowers the energies of all of the molecules in the sample. If there are 
several possible reactions that might occur and if they have different activation energies, we may be 
able to find a temperature where the population of molecules has only enough energy to surmount 
the lowest of the alternative energy barriers so that only one reaction occurs. The diazotization of 
aromatic amines is an example. The reaction involves treating the amine with nitrous acid (HONO) 
made from NaN0 2 and HC1. 






S0 2 CIF, SbF 5 

-120 °C 
liquid S0 2 



© 




NH, 



NaN0 2 , HCI 



H 2 0, 0-5 C 







© 



CI 



H 2 

room temperature 




OH 



At room temperature the diazonium salt decomposes to the phenol and cannot be used but 
at 0—5 °C it is stable and can be reacted with other nucleophiles in useful processes discussed in 
Chapter 23. 





NEt 2 



s-BuLi, THF 



-78 °C 



Other examples you have met 
involve lithiated organic molecules. 
NEt These are always prepared at low tem- 
peratures, often at -78 °C. The 
ortholithiation of aromatic amides was 
mentioned in Chapter 9. 
If the lithiation is carried out at °C, each molecule of lithiated amide attacks another molecule of 
unlithiated amide in the substitution reaction from Chapter 12. 








-78 °C is the convenient 
temperature of a bath of acetone 
with solid C0 2 dissolved in it. 



NEt 2 
OLi 



NEt, 



acid work-up 




332 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



The situation is more critical because of the behaviour of the solvent THF. This cyclic ether is a 
good solvent for lithiations because it is a good ligand for lithium and it remains liquid at -78 °C. But 
if lithiations are attempted at higher temperatures, THF also reacts with s-BuLi to give surprising by- 
products discussed in Chapter 35. 



o 



s-BuLi 



0°C 



-*- CH 2 



=CH, 



^ 



OLi 



We shall discuss this type of reaction- 
fragmentation — in Chapter 38. 



Solvents 

The nature of the solvent used in reactions often has a profound effect on how the reaction 
proceeds. Often we are limited in our choice of solvent by the solubilities of the reactants and 
products — this can also be to our advantage when trying to separate products, for example, in 
ether extractions. We have seen so far in this chapter that THF is a good solvent for lithiations 
because it coordinates to Li, that water is a good solvent for hydrolyses of carboxylic acids because it 
is a reagent and because it dissolves the carboxylate anion, and that alcohols are a good solvents in 
reactions such as transesterifications where mass action is needed to drive equilibria over towards 
products. 

But solvents can affect reactions more drastically; for example, the reaction below gives different 
products depending on the choice of solvent. 




PhCH 2 Br 

»- 




+ 




substituted naphthol 



an ether 



In water the product is almost all benzyl naphthol. However, in DMSO (dimethyl sulfoxide) the 
major product is the ether. In water the oxyanion is heavily solvated through hydrogen bonds to 
water molecules and the electrophile cannot push them aside to get close to 0~ (this is an entropy 
effect). DMSO cannot form hydrogen bonds as it has no OH bonds and does not solvate the oxy- 
anion, which is free to attack the electrophile. 





In terms of rates of reaction, where a charged intermediate is formed, a polar solvent will help to 
stabilize the charge by solvation. Some of this stabilization will already be present in the transition 
state and solvation will therefore lower the activation energy and speed up the reaction. Turning to a 
reaction not dealt with elsewhere in the book, an elimination of carbon dioxide, let us see how the 
rate constant varies with solvent. 



2 N 




0,N 




Solvents 



333 



These solvents may be divided into 
three groups — those in which the reaction 
is slower than in benzene, those in which it 
is faster, and, of course, benzene itself. The 
solvents in which the reaction goes rela- 
tively slowly all have something in com- 
mon — they have either O-H or N-H 
groups. Solvents of this kind are described 
as protic solvents, that is, they are capable 
of forming hydrogen bonds in solution 
(though none of these solvents is a good 
acid). Mechanistically, the important point 
is that these solvents solvate both cations 
and anions. The cations are solvated by use 
of the lone pairs on the oxygen or nitrogen; 
the anions via the hydrogens. 

We can illustrate this with a schematic 
drawing of the solvation of a salt (NaBr) by 
water. 



Rate of reaction in various solvents 

Solvent Rate 3 

H 2 0.0015 

MeOH 0.052 

HC0NH 2 0.15 

C6H6 1 

acetonitrile, CH 3 CN 600 

dimethyl sulfoxide, DMSO, (CH 3 ) 2 S0 2 100 

acetone, (CH 3 ) 2 C0 5 000 

dimethyl formamide, DMF, HC0N(CH 3 ) 2 7 700 

dimethylacetamide, CH 3 C0N(CH 3 ) 2 33 000 
hexamethyl phosphoramide, HMPA, [(CH 3 ) 2 N] 3 P0 150 000 
3 Relative to reaction in benzene. 



V H 
Y-nU- 

/ : 

H , 



b— h- 



0"^ 

I 

H 

i 
i 

-Br®--H— 

i. v h 

I 
o 



Solubilities of sodium bromide in protic solvents 



Solvent 

H 2 

MeOH 
EtOH 



Solubility, g/100 g of solvent 

90 

16 
6 



The solvents in which the reaction proceeds fastest also have something in common — they have 
an electronegative group (oxygen or nitrogen) but no O-H or N-H bonds. This class is known 
as polar aprotic solvents. Aprotic solvents can still solvate cations but they are unable to solvate 
anions. 

We can now understand the observed trend in the reaction. In the aprotic solvents, the positively 
charged counterion is solvated and, to some extent, separated from the anion. The anion itself is not 
solvated and hence is not stabilized; it can therefore react very easily. In protic solvents, such as 
water, the anion is stabilized by solvation and so is less reactive. We could represent this information 
on an energy level diagram (overleaf). The main effect of the solvent is on the energy of the starting 
material — good solvation lowers the energy of the starting material. 

The reaction in the aprotic solvent proceeds fastest because the activation energy for this reaction 
is smallest. This is not because the energy of the transition state is significantly different but because 
the energy of the starting material has been raised. You might wonder why the energy of the tran- 
sition state is not stabilized to the same extent as the starting material on changing from an 
aprotic solvent to a protic solvent. This is because the charge is spread over a number of atoms in 
the transition state and so it is not solvated to the same extent as the starting material, which 
has its negative charge localized on the one atom. This is an important point since, if the transition 
state were stabilized by the same amount as the starting materials, then the reaction would proceed 
just as quickly in the different solvents since they would then have the same activation energy 
barriers. 

When you meet the new reactions awaiting you in the rest of the book you should reflect that each 
is controlled by an energy difference. If it is an equilibrium, AG° must be favourable, if a kinetically 
controlled reaction, AG* must be favourable, and either of these could be dominated by enthalpy or 
entropy and could be modified by temperature control or by choice of solvent. 



334 



13 ■ Equilibria, rates, and mechanisms: summary of mechanistic principles 



transition 
state 



H* 



'Wo 



solvation 



t 




the energy of the transition state 
does not change significantly on 
solvation 



reaction profile for 
aprotic solvent 
reaction profile 
for protic solvent 



energy 



solvation of the starting 

material by the protic solvent 

lowers its energy 



2 N 



products 



2 N 




progrees of reaction 

Summary of mechanisms from Chapters 6-12 

We last discussed mechanisms in Chapter 5 where we introduced basic arrow-drawing. A lot has 
happened since then and this is a good opportunity to pull some strands together. You may like to be 
reminded: 

1 When molecules react together, one is the electrophile and one the nucleophile 

2 In most mechanisms electrons flow from an electron-rich to an electron-poor centre 

3 Charge is conserved in each step of a reaction 

These three considerations will help you draw the mechanism of a reaction that you have not 
previously met. 

Types of reaction arrows 

1 Simple reaction arrows showing a reaction goes from left to right or right to left 

NOH NOH 

II + NH 2 OH ^ II II -* II + NH 2 OH 



2 Equilibrium arrows showing extent and direction of equilibrium 

biased to 
about 50:50 f> the right 

II + EtOH _ II + H 2 II + EtNH 2 ^- * * II 



^OEt 



^OH 



© 
+ EtNH 3 

© 



Summary of mechanisms from Chapters 6-12 335 



3 Derealization or conjugation arrows showing two different ways to draw the same molecule. 
The two structures ('canonical forms' or 'resonance structures') must differ only in the position 
of electrons 

0° ^^ ^\ 

(7 ^-i delocalized 71 bonds r^ ?i 



delocalized anion 



Types of curly arrows 

1 The curly arrow should show clearly where the electrons come from and where they go to 

p H \f /"> O ho \/° 

^^^Jl ^ V is better than / ^ II S \J 

HO© R >^x R^S HO R ^S R^S 

2 If electrophilic attack on a Jt or bond leads to the bond being broken, the arrows should show 
clearly which atom bonds to the electrophile 

H® 

O ^* H ® ° © ®° /H 

p or MJ is better than K ^H when the product is 

R^^X R^^X R^^X R^^X 

3 Reactions of the carbonyl group are dominated by the breaking of the 7t bond. If you use this 
arrow first on an unfamiliar reaction of a carbonyl compound, you will probably find a 

reasonable mechanism 

addition (Chapter 6) substitution (Chapter 12) 

9^ is part of all the <>*) 0"*) H0 (A 

carbonyl reaction ( ~^ v sk. \\ / — ^JJ \/* — 

R^X in Chapters 6-12 CN © R > 1 \ H H Q R >S — R^\j 

Short cuts in drawing mechanisms 

1 The most important is the double-headed arrow on the carbonyl group used during a 
substitution reaction 

I P "\& \\ 

^OH + X ° iS ~™ HO©"^ — R^ ^R^OH + X& 

2 The symbol ±H + is shorthand for the gain and loss of a proton in the same step (usually involving 
N,0,orS) 




H 3 N 0© ± H© H 2 N OH rN ©J 2 O 0^ H J} X HzN 

V »~ V is equivalent to: X ° T H "VK/ ^ W 



Nucleophilic substitution at C=0 with 
loss of carbonyl oxygen 



14 



Connections 


Building on: 


Arriving at: 


Looking forward to: 


• Nucleophilic attack on carbonyl 


• Replacement of carbonyl oxygen 


• Protecting groups ch24 


groups ch6 


• Acetal formation 


• Synthesis in action ch25 


• Nucleophilic substitution at carbonyl 
groups chl2 

• Acidity and pK a ch8 

• Rate and pH chl3 


• Imine formation 


• Acylation of enolates ch28 


• Stable and unstable imines 

• Reductive amination 

• The Strecker and Wittig reactions 


• Synthesis of amino acids ch49 

• Synthesis of alkenes ch31 

• Stereochemistry chl6 






• Asymmetric synthesis ch45 



Introduction 

Nucleophiles add to carbonyl groups to give compounds in which the trigonal carbon atom of the 
carbonyl group has become tetrahedral. 

Ah® 

r U OH 

nucleophilic addition 
to a carbonyl group 




Nu 



Nu 



In Chapter 12 you saw that these compounds are not always stable: if the starting material contains a 
leaving group, the addition product is a tetrahedral intermediate, which collapses with loss of the leaving 
group to give back the carbonyl group, with overall substitution of the leaving group by the nucleophile. 



nucleophilic substitution 
at a carbonyl group 




P 






Nu 



In this chapter, you will meet more substitution reactions of a different type. Instead of losing a 
leaving group, the carbonyl group loses its oxygen atom. Here are three examples: the carbonyl oxy- 
gen atom has been replaced by an atom of O, a nitrogen atom, and two atoms of oxygen. Notice 
too the acid catalyst — we shall see shortly why it is required. 



H 2 18 



nucleophilic substitution 
at a carbonyl group 
with loss of 
carbonyl oxygen 



16 n I8n 

O C at H O 

A^A 



+ H, 16 



^CHO 




H,0 



OH 



340 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



You have, in fact, already met some reactions in which the carbonyl oxygen atom can be lost, but 
you probably didn't notice at the time. The equilibrium between an aldehyde or ketone and its 
hydrate (p. 000) is one such reaction. 



OPh 




In Chapter 13 we saw this way of 
making a reaction go faster by raising 
the energy of the starting material. We 
also saw that the position of an 
equilibrium can be altered by using a 
large excess of one of the reagents. 
This is often called a mass action 
effect. 



H,0 + 



A. 



HO OH 



When the hydrate reverts to starting materials, either of its two oxygen atoms must leave: one 
came from the water and one from the carbonyl group, so 50% of the time the oxygen atom that 
belonged to the carbonyl group will be lost. Usually, this is of no consequence, but it can be useful. 
For example, in 1968 some chemists studying the reactions that take place inside mass spectrometers 
needed to label the carbonyl oxygen atom of this ketone with the isotope O. 

By stirring the 'normal' O compound with a large excess of isotopically labelled water, H2 O, 
for a few hours in the presence of a drop of acid they were able to make the required labelled com- 
pound. Without the acid catalyst, the exchange is very slow. Acid catalysis speeds the reaction up by 
making the carbonyl group more electrophilic so that equilibrium is reached more quickly. The 
equilibrium is controlled by mass action — O is in large excess. 



OPh 




OPh 



OPh 



OPh 





We need now to discuss hemiacetals though you may well wonder why - they retain the carbonyl 
oxygen and they are unstable. We need to discuss them as a preliminary to the much more important 
acetals. Hemiacetals are halfway to acetals. 



Aldehydes can react with alcohols to form hemiacetals 

When acetaldehyde is dissolved in methanol, a reaction takes place: we know this because the IR 
spectrum of the mixture shows that a new compound has been formed. However, isolating the prod- 
uct is impossible: it decomposes back to acetaldehyde and methanol. 



M 



MeOH 



IR: 


no peak in carbonyl 


region 1600-1800 


strong OH stretch 


3000-3500 



attempt to purify 



M 



MeOH 



The product is in fact a hemiacetal. Like hydrates, most hemiacetals are unstable with respect to 
their parent aldehydes and alcohols: for example, the equilibrium constant for reaction of acetalde- 
hyde with simple alcohols is about 0.5 as we saw in Chapter 13. 



MeO -OH 

rvwr^H 

hemiacetal 



K=~0.5 



+ MeOH — 



Me^ "H 

aldehyde 



MeO. ,OH 
MC^^H 



hemiacetal 



Aldehydes can react with alcohols to form hemiacetals 



341 



This equilibrium constant K is defined as 
[hemiacetal] 



K 



[aldehyde][MeOH] 



So by making [MeOH] very large (using it as the solvent, for example) we can turn most of the 
aldehyde into the hemiacetal. However, if we try and purify the hemiacetal by removing the 
methanol, more hemiacetal keeps decomposing to maintain the equilibrium constant. That is why 
we can never isolate such hemiacetals in a pure form. 

Only a few hemiacetals are stable 

Like their hydrates, the hemiacetals of most ketones (sometimes called hemiketals) are even less sta- 
ble than those of aldehydes. On the other hand, some hemiacetals of aldehydes bearing electron- 
withdrawing groups, and those of cyclopropanones, are stable, just like the hydrates of the same 
molecules. 



These are more 'mass action' effects 
like the 18 exchange we have just 
discussed. 



We discussed the reasons for this in 
Chapter 6. 



Br 3 C 



EtOH 



OEt 



Br 3 (T^0H 



a stable hemiacetal 



Hemiacetals that can be formed by intramolecular cyclization of an alcohol on to an aldehyde are 
also often stable, especially if a five- or six-membered ring is formed. You met this in Chapter 6 — 
many sugars (for example, glucose) are cyclic hemiacetals, and exist in solution as a mixture of open- 
chain and cyclic forms. 



glucose: 



open-chain form 
OH OH 



,CHO 



OH 

0.003% 



OH 



cyclic form 




Why are cyclic hemiacetals stable? 

Part of the reason for the stability of cyclic 
hemiacetals concerns entropy. Formation of an 
acyclic acetal involves a decrease in entropy (AS 
negative) because two molecules are consumed for 
every one produced . This is not the case for 
formation of a cyclic hemiacetal. Since AG" = AH°- 
7AS°, a reaction with a negative AS° tends to have a 
more positive AG"; in other words, it is less 
favourable. 

Another way to view the situation is to consider the 
rates of the forward and reverse processes. We can 



measure the stability of acyclic hemiacetal by the 
equilibrium constant Kforthe ring-opening reaction: 
a large Kmeans lots of ring-opened product, and 
therefore an unstable hemiacetal, and a small K 
means lots of ring-closed product: a stable 
hemiacetal. After reading Chapter 13 you should 
appreciate that an equilibrium constant is simply the 



equilibrium constant 
"forward 



HO. P 



rate of the forward reaction divided by the rate of the 
reverse reaction. So, for a stable hemiacetal, we 
need a fast hemiacetal-forming reaction. And when 
the hemiacetal is cyclic that is just what we do have: 
the reaction is intramolecular and the nucleophilic 
OH group is always held close to the carbonyl group, 
ready to attack. 



"reverse 
fast for cyclic hemiacetals 




Acid or base catalysts increase the rate of equilibration of hemiacetals with their 
aldehyde and alcohol parents 

Acyclic hemiacetals form relatively slowly from an aldehyde or ketone plus an alcohol, but their rate 
of formation is greatly increased either by acid or by base. As you would expect, after Chapters 12 
and 13, acid catalysts work by increasing the electrophilicity of the carbonyl group. 



342 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



acid-catalysed hemiacetal formation 



© 






Me 



© 
OH 



M 



HoCp£ 



Me 



HO OMe 



Me^ 



hemiacetal 



M 



S 



Base catalysts, on the other hand, work by increasing the nucleophilicity of the alcohol by remov- 
ing the OH proton before it attacks the C=0 group. In both cases the energy of the starting materials 
is raised: in the acid-catalysed reaction the aldehyde is destabilized by protonation and in the base- 
catalysed reaction the alcohol is destabilized by deprotonation. 



base-catalysed hemiacetal formation 



J£\ base makes alcohol 
N\er H more nucleophilic 



base-* 



Me' 






Me' 



© 

D 0- 

X 



-L-ha 



base 



/Me 

o <r 



y 



Me 



MtT^H 



hemiacetal 



You can see why hemiacetals are unstable: they are essentially tetrahedral intermediates contain- 
ing a leaving group and, just as acid or base catalyses the formation of hemiacetals, acid or base also 
catalyses their decomposition back to starting aldehyde or ketone and alcohol. That's why the title 
of this section indicated that acid or base catalysts increase the rate of equilibration of hemiacetals 
with their aldehyde and alcohol components — the catalysts do not change the position of that 
equilibrium! 



acid-catalysed hemiacetal decomposition 

/^H© 
HO OMe 



Me^^H 



Me^^H 



Me 



protonation makes the alcohol a 
better leaving group 

- MeOH + 



© 



tf 



base-catalysed hemiacetal decomposition 
JVIe 



base 



.JX * 



M 



H 



Me-^^H 



M 



deprotonation forces the 
alcohol to leave as alkoxide 

Me 



M 



M 



Meff 



© 



To summarize 

Hemiacetal formation and decomposition are catalysed by acid or base. 
ROH HO OR 



catalysed by 
acid or base 



K ^R 

hemiacetal 



Acetals are formed from aldehydes or ketones plus alcohols in 
the presence of acid 

We said that a solution of acetaldehyde in methanol contains a new compound: a hemiacetal. We've 
also said that the rate of formation of hemiacetals is increased by adding an acid (or a base) catalyst 
to an alcohol plus aldehyde mixture. But, if we add catalytic acid to our acetaldehyde-methanol 



Acetals are formed from aldehydes or ketones plus alcohols in the presence of acid 343 

mixture, we find not only that the rate of reaction of the acetaldehyde with the methanol increases, 
but also that a different product is formed. This product is an acetal. 

MeOH Me0 s / 0H MeOH Me0 - 0Me 



""■«< "H H+cata|yst Me- H H+ cata|yst Me- H 

hemiacetal intermediate acetal 

In the presence of acid (but not base!) hemiacetals can undergo an elimination reaction (different 
from the one that just gives back aldehyde plus alcohol), losing the oxygen atom that once belonged 
to the parent aldehyde's carbonyl group. The stages are: 

1 Protonation of the hydroxyl group of the hemiacetal 

2 Loss of water by elimination. This elimination leads to an unstable and highly reactive oxonium 
ion 

3 Addition of methanol to the oxonium ion (breaking the 7t bond and not the <T bond, of course) 

4 Loss of a proton to give the acetal 



acid-catalysed acetal formation from hemiacetal 



unstable oxonium ion 



H ® u \- ™. u § m«„ J±k^ Me 

HO. .OMe H 3°\V / r^ 



S ^H Me^ ^H Me^N H Me X^™* Ma ^ 

\-.. H 



+ H© 
OMe 



MeOH 
Oxonium ions 

Oxonium ions have three bonds to a positively charged oxygen atom. All three bonds can be a bonds as in H 3 + or Meerwein'ssalt, trimethyloxoniumfluoroborate, 
a stable (though reactive) compound described in Chapter 21, or one bond can be a jt bond as in the acetal intermediate. The term 'oxonium ion' describes either 
of these structures. They are like alkylated ethers or O-alkylated carbonyl compounds. 



,R 2 RV^R 2 Me \®/ Me O ®0^ R3 

ii Tm I BF ? 

alkylation U ^ e 



i-^„2 alk y |ation D i/\ E 



oxonium ion Meerwein's salt oxonium ion 

trimethyloxonium tetrafluoroborate 

Just as protonated carbonyl groups are much more electrophilic than unprotonated ones, these 
oxonium ions are powerful electrophiles. They can react rapidly with a second molecule of alcohol to 
form new, stable compounds known as acetals. An oxonium ion was also an intermediate in the for- 
mation of hemiacetals in acid solution. Before reading any further, it would be worthwhile to write 
out the whole mechanism of acetal formation from aldehyde or ketone plus alcohol through the 
hemiacetal to the acetal, preferably without looking at the fragments of mechanism above, or the 
answer below. 

• Formation of acetals and hemiacetals 

Hemiacetal formation is catalysed by acid or base, but acetal formation is possible 
only with an acid catalyst because an OH group must be made into a good leaving 
group. 

ROH HO OR ROH RO OR 

11 . V . V 

R R R' R R R 

" " catalysed by " " catalysed by " " 

acid or base acid only 

ketone hemiacetal acetal 

When you look at our version of this complete mechanism you should notice a remarkable degree 
of similarity in the two halves. The reaction starts with a protonation on carbonyl oxygen and, when 



344 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



you get to the temporary haven of the hemiacetal, you start again with protonation of that same oxy- 
gen. Each half goes through an oxonium ion and each oxonium ion adds the alcohol. The last step in 
the formation of both the acetal and the hemiacetal is the loss of a proton from the recently added 
alcohol. 

This is about as complex a mechanism as you have seen and it will help you to recall it if you see it 
in two halves, each very similar to the other. First, form the hemiacetal by adding an alcohol to the 
C=0 Jt bond; then lose the OH group by breaking what was the C=0 bond to form an oxonium 
ion and add a second alcohol to form the acetal. From your complete mechanism you should also be 
able to verify that acetal formation is indeed catalytic in acid. 



acid-catalysed acetal formation 



excess alcohol, removal of water 



6 



"l^ 



oxonium ions 
attacked by alcohol 




deprotonation of adduct 

H 

f\/ R 
HO.% 



■X 



OR 3 



R 3 OH 



R ^N< 3 

R 2 
H 



!. /R 
,0 



hemiacetal 
intermediate 



OR 3 



R 2 

acetal 



^OR 3 



+ H© 



excess water 



Remember the oxonium ion! 

When you wrote out your mechanism for acetal formation, we hope you didn't miss out the oxonium ion! It's easy to do 
so, but the mechanism most definitely does not go via a direct displacement of water by alcohol. 



J^ 



R l/V R 2 



'^R 2 ^ _ R 1 



R 1 ^ V"R : 

R 3 0H 

oxonium ion 



0R J 



R 2 O 



© 




D 



R 3 0H 



INCORRECT STEP 



If you wonder how we know this, consult a specialized book on organic reaction mechanisms. After you have read 
Chapter 17 in this book, you will be able to spot that this substitution step goes via an SnI and not an Sn2 mechanism. 






Making acetals 

Just as with the ester formation and hydrolysis reactions we discussed in Chapters 12 and 13, every 
step in the formation of an acetal is reversible. To make acetals, therefore, we must use an excess of 
alcohol or remove the water from the reaction mixture as it forms, by distillation, for example. 





K~ 1 



+ 1 x EtOH z^ 



^N>Et ^N 



K-0.01 Ettt OEt 
+ 2 x EtOH , ">/ + H 2 

M«T ^H 



In fact, acetal formation is even more difficult than ester formation: while the equilibrium con- 
stant for acid-catalysed formation of ester from carboxylic acid plus alcohol is usually about 1, for 



Acetals are formed from aldehydes or ketones plus alcohols in the presence of acid 



345 



acetal formation from aldehyde and ethanol (shown above), the equilibrium constant is K= 0.0125. 
For ketones, the value is even lower: in fact, it is often very difficult to get the acetals of ketones (these 
used to be called ketals) to form unless they are cyclic (we consider cyclic acetals later in the chapter). 
However, there are several techniques that can be used to prevent the water produced in the reaction 
from hydrolysing the product. 



acetaldehyde 
present ■-■ 

in excess 



MeCHO 



toluenesulfonic 

acid catalyst 

TsOH 

heat, 12 h 




50% yield of acetal 



In these two examples, with the more reactive aldehyde, it was sufficient just to have an excess of 

one of the reagents (acetaldehyde) to drive the reaction to completion. Dry HC1 gas can work too. In 

the second example, with a less reactive ketone, molecular sieves (zeolite) were used to remove water 

from the reaction as it proceeded. 

OMe 
Me^ ^CHO MeOH 



Me 



pass dry HCI gas 
2 min, 20 °C 




OMe 



A 



R 

60% yield 



catalytic TsOH 

molecular sieves 

0'C, 2h 



X 



62% yield 



para-Toluenesulfonic acid is 
commonly used to catalyse 
reactions of this sort. It is a stable 
solid, yet is as strong an acid as 
sulfuric acid. It is widely available 
and cheap because it is produced 
as a by-product in the synthesis of 
saccharin (for more details, see 
Chapter 22). <x n 

*^ 



c-toluenesulfonic acid 





Overcoming entropy: orthoesters 

We have already mentioned that one of the factors that makes acyclic hemiacetals unstable is the 
unfavourable decrease in entropy when two molecules of starting material (aldehyde or ketone plus 
alcohol) become one of product. The same is true for acetal formation, when three molecules of 
starting material (aldehyde or ketone plus 2 x alcohol) become two of product (acetal plus H 2 0). We 
can improve matters if we tie the two alcohol molecules together in a diol and make a cyclic acetal: 
we discuss cyclic acetals in the next section. Alternatively, we can use an orthoester as a source of 
alcohol. Orthoesters can be viewed as the 'acetals of esters' or as the triesters of the unknown 
'orthoacids' — the hydrates of carboxylic acids. They are hydrolysed by water, catalysed by acid, to 
ester + 2 x alcohol. 



orthoacids don't exist 



orthoesters 



OH 



M 



OH 
OH 



OEt 

r> Et 

Me^^^OEt 



OMe 
I^OMe 

H'^^OMe 



H,0 



u 

X 



+ 2 MeOH 



OMe 



orthoacetic acid 



triethyl orthoacetate trimethyl orthoformate 



Here is the mechanism for the hydrolysis — you should be feeling quite familiar with this sort of 
thing by now. 

- Me TbH 



OMe 
I^OMe 

H^^OMe 



© 



VJx>Me 
H^Me 
H 




H® C? Me nM. HbHVfo ±H @ M/ 0Me -H® 



OMe 



/V0 
H (JOMe 




H' ^OMe 






Ketones or aldehydes can undergo acetal exchange with orthoesters. The mechanism starts off as 
if the orthoester is going to hydrolyse but the alcohol released adds to the ketone and acetal forma- 
tion begins. The water produced is taken out of the equilibrium by hydrolysis of the orthoester. 



OMe 
T>Me + 

H^^OMe 

trimethyl orthoformate 




MeOH, H + cat. 



20 °C, 15 min 



X 



H ^OMe 

methyl formate 




346 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



Acetals hydrolyse only in the presence of acid 

Just as acetal formation requires acid catalysis, acetals can be hydrolysed only by using an acid cata- 
lyst. With aqueous acid, the hydrolysis of acyclic acetals is very easy. Our examples are the two acetals 
we made earlier. 




3% HCI, H 2 



30 min 



MeCHO + 2 BuOH 



OMe 




2 M H 2 S0 4 



OMe 



CHO 



+ 2 MeOH 



We hope you didn't 
make the mistake of 
missing out the 
oxonium ion step! 



• Acetal hydrolysis 

Acetals can be hydrolysed in acid but are stable to base. 

We won't go through the mechanism again — you've already seen it as the reverse of acetal formation 
(and you have a hint of it in the orthoester hydrolysis just discussed), but the fact that acetals are stable to 
base is really a very important point, which we will use on p. 000 and capitalize on further in Chapter 24. 

Cyclic acetals are more stable towards hydrolysis than acyclic ones 

Of course you want us to prove it: well — 



MeO 



Me 




CF 3 C0 2 H 
H 2 

*- 

CHCI 3 (solvent) 
"C, 1 h 



OH 




96% yield 



The acetals you have met so far were formed by reaction of two molecules of alcohol with one of 
carbonyl compound. Cyclic acetals, formed by reaction of a single molecule of a diol, a compound 
containing two hydroxyl groups, are also important. When the diol is ethylene glycol (as in this 
example) the five-membered cyclic acetal is known as a dioxolane. 

ethylene glycol 



cat. TsOH, heat, 
remove water by distillation 



o_ JO 



78% yield 



Before looking at the answer below, try to write a mechanism for this reaction. If you need it, use 
the mechanism we gave for the formation of acyclic acetals. 

acid-catalysed dioxolane formation 



,0H 



H' 



© 



H' 



© 



,OH 




HO 



H 

<4 



OH ^ 



© 



c 

HO ( 



hemiacetal intermediate 



,OH 




fv r^© 

1 ^ 0<V H 




o_ JO 



oxonium ion 



the dioxolane 



Acetals are formed from aldehydes or ketones plus alcohols in the presence of acid 



347 



Cyclic acetals like this are more resistant to hydrolysis than acyclic ones and easier to make — they 
form quite readily even from ketones. Again, we have entropic factors to thank for their stability. For 
the formation of a cyclic acetal, two molecules go in (ketone plus diol) and two molecules come out 
(acetal plus water), so the usually unfavourable AS° factor is no longer against us. And, as for hemi- 
acetals (see the explanation above), equilibrium tends to lie to the acetal side because the intramole- 
cular ring-closing reaction is fast. 

Water is still generated, and needs to be got rid of: in the example above you can see that water 
was distilled out of the reaction mixture. This is possible with these diols because they have a boil- 
ing point above that of water (the boiling point of ethylene glycol is 197 C C). You can't distil water 
from a reaction mixture containing methanol or ethanol, because the alcohols distil too! One very 
useful piece of equipment for removing water from reaction mixtures containing only reagents that 
boil at higher temperatures than water is called a Dean Stark head: there is a picture of this in 
Chapter 13. 



Modifying reactivity using acetals 

Why are acetals so important? Well, they're important to both nature and chemists because many 
carbohydrates are acetals or hemiacetals (see the box below). One important use that chemists have 
put them to is as protecting groups. 

One important synthesis of the steroid class of compounds (about which more later) requires a 
Grignard reagent with this structure. 




* 




We shall discuss protecting groups in much 
more detail in Chapter 24. 



unstable structure 
- impossible to make 



Yet this compound cannot exist: it would react with itself. Instead, this Grignard reagent is used, 
made from the same bromoketone, but with an acetal-forming step. 



Don't be confused by this 
statement! Acetal formation and 
hydrolysis are invariably carried 
out under thermodynamic 
control — what we mean here is 
that the equilibrium constant for 
acetal hydrolysis, which is a 
measure of rate of hydrolysis 
divided by rate of formation, turns 
out to be small because the rate 
of formation is large. 

► Dean Stark head 

When a mixture of toluene and 
water boils, the vapour produced 
is a constant ratio mixture of 
toluene vapour and water vapour 
known as an azeotrope. If this 
mixture is condensed, the liquid 
toluene and water, being 
immiscible, separate out into two 
layers with the water below. By 
using a Dean Stark apparatus, or 
Dean Stark head, the toluene 
layer can be returned to the 
reaction mixture while the water is 
removed. Reactions requiring 
removal of water by distillation are 
therefore often carried out in 
refluxing toluene or benzene 
under a Dean Stark head. 



Br 



H + cat. 




Mg, Et 2 




Acetals, as we stressed, are stable to base, and to basic nucleophiles such as Grignard reagents, so 
we no longer have a reactivity problem. Once the Grignard reagent has reacted with an electrophile, 
the ketone can be recovered by hydrolysing the acetal in dilute acid. The acetal is functioning here as 



Acetals in nature 

We showed you glucose as an example of a stable, cyclic hemiacetal. Glucose can, in fact, react with itself to form an acetal known as maltose. 





acetal 



maltose 




Maltose isadisaccharide (made of two sugar units) produced by the enzymatic hydrolysis of starch or cellulose, which are themselves polyacetals made up of a 
string of glucose units. 



348 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



a protecting group because it protects the ketone from attack by the Grignard reagent. Protecting 
groups are extremely important in organic synthesis, and we will return to them in Chapter 24. 



Amines react with carbonyl compounds 

The ketone carbonyl group of pyruvic acid (or 2-oxopropanoic acid) has a stretching frequency 
of a typical ketone, 1710 cnT . When hydroxylamine is added to a solution of pyruvic acid, this 
stretching frequency slowly disappears. Later, a new IR absorption appears at 1400 cnT . What 
happens? 



IR 1710 cm- 



absorption 



M 



9 H 

-4 + i 



-»- ? 



2" M OH 

hydroxylamine 



IR 1400 cm- 




time 

Well, you saw a diagram like this in the last chapter when we were discussing kinetic and thermo- 
dynamic products (p. 000) and you can probably also apply something of what you now know about 
the reactivity of carbonyl compounds towards nucleophiles to work out what is happening in this 
reaction between a carbonyl compound and an amine. The hydroxylamine first adds to the ketone to 
form an unstable intermediate such as a hemiacetal. 
intermediate formation 



© 
NHOH 



Me 




±H 



,© 



HO. NHOH 



„/\r 



Ph^ ^CO? 

intermediate 



Notice that it is the more nucleophilic nitrogen atom, and not the oxygen atom, of hydroxy- 
lamine that adds to the carbonyl group. Like hemiacetals, these intermediates are unstable and can 
decompose by loss of water. The product is known as an oxime and it is this compound, with its 
C=N double bond, that is responsible for the IR absorption at 1400 cnT . 
dehydration of the intermediate to give oxime 



H® H 

Oo 



M«r> 



NHOH 



CO® 




H^OH 



,OH 



M«r> 



CO® 



II"* IR 1400 cm" 



XO® 



intermediate oxime 

We know that the oxime is formed via an intermediate because the 1400 cnT absorption hardly 
appears until after the 1710 cm absorption has almost completely gone. We also know something 
must be there because, by IR, everything has disappeared. There must really be another curve to 
show the formation and the decay of the intermediate, the hemiacetal, just like the one in the last 
chapter (p. 000). The only difference is that the intermediate has no double bond to give an IR 
absorbance. We come back to oximes later in the chapter. 



Amines react with carbonyl compounds 



349 



HO. NHOH 



concentration 



U 

Mcr^co 

pyruvate 




time 



N 



s" 



N 



Imines are the nitrogen analogues of carbonyl compounds 

In fact, the oxime formed from a ketone and hydroxylamine is just a special example of an 

imine. , J*+^ , , J**+. 

R R R 

Imines are formed when any primary amine reacts with an aldehyde or a ketone under appro- 

. an imine an oxim 

priate conditions: for example, cyclohexylamine and benzaldehyde. 



OH 




+ H 2 



You shouldn't need us to tell you the mechanism of this reaction: even without looking at 
the mechanism we gave for the formation of the oxime it should come as no surprise to you by 
now. First, the amine attacks the aldehyde and the intermediate is formed. Dehydration gives the 
imine. 



Vo 



this step rate-determining below pH4 

A 






^ 



r 

OH 







r 



acid catalyst needed - slow above pH 6 

A v 



J^\^T\,/ Ph ^ ^\©/ ph — — ^L /Ph — — SItn Ph — — ^^©/ ph — — /^ / Ph 
PI^^H H 2N ^Ph^^N^ ^Ph^^N^ ^Ph^Ni^ Ph if ^Ph^ ^N 



A* 



/\ 
H H 



H 

hemiaminal 



1 



# Imine formation requires acid catalysis. 

Notice that an acid catalyst is normally added for imine formation. Without an acid catalyst, the 
reaction is very slow, though in some cases it may still take place (oximes, for example, will form 
without acid catalysis, but form much faster with it). It's important to notice that acid is not needed 
for the addition step in the mechanism (indeed, protonation of the amine means that this step is very 
slow in strong acid), but is needed for the elimination of water later on in the reaction. Imine forma- 
tion is in fact fastest at about pH 4-6: at lower pH, too much amine is protonated and the rate of the 
first step is slow; above this pH the proton concentration is too low to allow protonation of the OH 
leaving group in the dehydration step. Imine formation is like a biological reaction: it is fastest near 
neutrality. 



350 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



Because it is made from an 
unsymmetrical ketone, this imine 
can exist as a mixture of Eand Z 
isomers, just like an alkene. 
When it is formed by this method, 
the ratio obtained is 8:1 £;Z. 
Unlike the geometrical isomers of 
alkenes, however, those of an 
imine are usually unstable and 
interconvert quite rapidly at room 
temperature. The geometrical 
isomers of oximes, on the other 
hand, are stable and can even be 
separated. 



log rate 



variation of rate of imine formation with pH 



\\ RN " 2 - \\ 

X„ — - „ A D ♦ H 2 



rate-determinin, 
addition 



rate-determining 
dehydration 




-5-6 



Either side of pH 5-6 the reaction goes more slowly. This is a sign of a change in rate- 
determining step. Where there is a choice between two rate-determining steps, the slower of the 
two determines the overall rate of the reaction. In the last chapter we saw that ester hydrolysis 
was a typical example of an organic reaction showing acid and base catalysis. It has a mini- 
mum rate at about neutrality showing that the mechanism must change. Where there is a choice 
of mechanism, the faster of the two operates. The contrast between the two is obvious from the 
diagrams. 



log rate 



variation of rate of ester hydrolysis with pH 



u u 

X — -X 

^N)R H 2 /\ H 



ROH 




> Multistep reaction rates 

The overall rate of a multistep reaction is decided by: 

• The faster of two available mechanisms 

• The slower of two rate-determining steps 

Imines are usually unstable and are easily hydrolysed 

Like acetals, imines are unstable with respect to their parent carbonyl compound and amine, and 
must be formed by a method that allows removal of water from the reaction mixture. 

Me 



I 
PIT^Me 



Me 



H 2 N 



cat. H 



© 



Ph 



benzene reflux 
Dean Stark 



N ^Ph 

X 

PIT^Me 



imine 
72% yield 



Imines are formed from aldehydes or ketones with most primary amines. In general, they are only 
stable enough to isolate if either the C or N of the imine double bond bears an aromatic substituent. 
Imines formed from ammonia are unstable, but can be detected in solution. CH2=NH2, for exam- 
ple, decomposes at temperatures above -80 °C, but PhCH=NH is detectable by UV spectroscopy in a 
mixture of benzaldehyde and ammonia in methanol. 




CHO 



NH, 




NH 



H,0 



Imines are readily hydrolysed back to carbonyl compound and amine by aqueous acid — in fact, 



Amines react with carbonyl compounds 



351 



except for the particularly stable special cases we discuss on p. 000, most can be hydrolysed by water 
without acid or base catalysis. 




PhMgBr 



CN 

nitrile 




N©MgBr© 




H^ H 2 

— *~ X. ,Ph — v. » 



T 



NH 2 

unstable imine 



NH 3 




You have, in fact, already met an imine hydrolysis: at the end of Chapter 12 we talked about the 
addition of Grignard reagents to nitriles. The product is an imine that hydrolyses in acid solution to 
ketone plus ammonia. 





H 2 N C0H 





Ph 



H 3 l\iy OH 





Ph 



Some imines are stable 

Imines in which the nitrogen atom carries an electronegative group are usually stable: examples 
include oximes, hydrazones, and semicarbazones. 



phenyl- 
hydrazone 



.OH 



Ph 

I 
,NH 



NH 2 OH 



(hydroxylamine) 



PhNHNH 3 



(phenylhydrazine) 



ketone 



0^ ,NH 2 
A<T 

I 
,NH 

NH;f 



(semicarbazide) 



.NH 
N 



semicarbazone 



These compounds are more stable than imines because the electronegative sub- 
stituent can participate in derealization of the imine double bond. Derealization 
decreases the 8+ charge on the carbon atom of the imine double bond and raises the 
energy of the LUMO, making it less susceptible to nucleophilic attack. 

Oximes, hydrazones, and semicarbazones require acid or base catalysis to be hydrolysed 



HO^ 




H 2 S0 4 



H 2 




70% yield 



•Q> H 




Historical note 

Because the hydrazone and semicarbazone derivatives of these three isomeric five-carbon ketones are all similar, 

carbonyl compounds are often stable, crystalline solids, and before the days of NMR spectroscopy it would have 

they used to be used to confirm the supposed identity of been hard to distinguish between them, 
aldehydes and ketones. For example, the boiling points of 




b.p. 102 °C 



b.p. 102 °C 



b.p. 106 °C 



352 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



Historical note (continued) 

Their semicarbazones and 2,4-dinitrophenylhydrazones, 
on the other hand, all differ in their melting points. By 
making these derivatives of the ketones, identification 
was made much easier. Of course, all of this has been 
totally superseded by NMR! Howeverthese crystalline 
derivatives are still useful in the purification of volatile 



aldehydes and ketones and in solving structures t 
crystallography. 



>-.NH 2 0^NH 2 




m.p. 112 °C 



m.p. 139 °C 



m.p. 157 °C 



m.p. 143 °C 



m.p. 156 °C 



m.p. 125 °C 



Iminium ions and oxonium ions 



Let's return to the mechanism of imine formation, and compare it for a moment with that of 
acetal formation. The only difference to begin with is that there is no need for acid catalysis for 
the addition of the amine but there is need for acid catalysis in the addition of the alcohol, a much 
weaker nucleophile. 



acid-catalysed imine formation 



51 



© 



H 

\©/R 3 



,3/ 



NH 2 



Q o n: 

y h 



r 

HO NHR 3 

R^^R 2 

intermediate 



h^Tnhr3 



®/R 3 

HINT 



acid-catalysed acetal formation 

A® 

© 

'OH 




R 1 ^ 



f ® H f 

! XT C\/ R3 ' 

( R^^R 2 






,3/ 



OH 



HO OR 3 

R^^R 2 

hemiacetal 
intermediate 



H^TOR 3 



iminium ion 

R 3 
© / R 

x 

R 1^\ R 2 



oxonium ion 



Up to this point, the two mechanisms follow a very similar path, with clear analogy be- 
tween the intermediate and hemiacetal and the iminium and oxonium ion. Here, though, they 
diverge, because the iminium ion carries a proton, which the oxonium ion doesn't have. The 
iminium ion therefore acts as an acid, losing a proton to become the imine. The oxonium ion, 
on the other hand, acts as an electrophile, adding another molecule of alcohol to become the 
acetal. 



iminium ion 



H ®^R 3 



oxonium ion 



imine R 3 \ ./ H 



acetal 



As you might guess, however, iminium ions can be persuaded to act as electrophiles, just 
like oxonium ions, provided a suitable nucleophile is present. We will spend the next few pages 
considering reactions in which an iminium ion acts as an electrophile. First, though, we will look 
at a reaction in which the iminium ion cannot lose an N-H proton because it has none. 



Amines react with carbonyl compounds 



353 



Secondary amines react with carbonyl compounds to form enamines ■ 

Pyrrolidine, a secondary amine, reacts with isobutyraldehyde, under the sort of conditions you [cfc n d™ e we n bOTd^a C nTam?n S e ePe 



would use to make an imine, to give an enamine. 




TsOH catalyst 



benzene, heat 
-H 2 (Dean Stark) 




enamine 
94-95% yield 



The mechanism consists of the same steps as those that take place when imines form from pri- 
mary amines, up to formation of the iminium ion. This iminium ion has no N-H proton to lose, so 
it loses one of the C-H protons next to the C=N to give the enamine. Enamines, like imines, are un- 
stable to aqueous acid. We shall return to them in Chapter 21. 









only proton iminium ion 
can lose is this one 



^^"^ 



secondary amine 
(pyrrolidine) 



> Imines and enamines 

• Imines are formed from aldehydes or ketones with primary amines 

• Enamines are formed from aldehydes or ketones with secondary amines 

• Both require acid catalysis and removal of water 

primary amine secondary amine 



A 



H,N' 



cat. H + 
-H,0 



A 



H 



cat. H + 
-H,0 



A 



Enamines of primary amines, or even of ammonia, also exist, but only in equilibrium with an 
imine isomer. The interconversion between imine and enamine is the nitrogen analogue of enoliza- 
tion, which is discussed in detail in Chapter 21. 





CHO 



H,N 






Iminium ions can react as electrophilic intermediates 

We made the point above that the difference in reactivity between an iminium ion and an oxonium 
ion is that an iminium ion can lose H + and form an imine or an enamine, while an oxonium ion 
reacts as an electrophile. Iminium ions can, however, react as electrophiles provided suitable nucleo- 
philes are present. In fact, they are very good electrophiles, and are significantly more reactive than 



354 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



carbonyl compounds. For example, iminium ions are reduced rapidly by the mild reducing agent 
sodium cyanoborohydride (NaCNBH 3 ), while carbonyl compounds are not. 



Sodium cyanoborohydride 
contains the cyanoborohydride 
anion, whose structure is 

CN 

M 

H 

It is a 'toned down' version of 
sodium borohydride — the 
electron-withdrawing cyano group 
decreases the ease with which 
hydride is transferred. 




Me^©^Me 



NaCNBHo 



pH 6 



no reaction 




NaCNBH 3 



pH 6 




An alternative to Na(CN)BH3 is NaBH(OAc)3 (sodium triacetoxy-borohydride) — somewhat 
safer because strong acid can release HCN from Na(CN)BH3. 

Amines from imines: reductive amination 

A useful way of making amines is by reduction of imines (or iminium ions). This overall process, 
from carbonyl compound to amine, is called reductive amination. This is, in fact, one of the few suc- 
cessful ways, and the best way, of making secondary amines. This should be your first choice in 
amine synthesis. 



F^NHa 






.© 



IN 



2 x [H] 



HN' 



H 

secondary amine 



This can be done in two steps, provided the intermediate is stable, but, because the instability of 
many imines makes them hard to isolate, the most convenient way of doing it is to form and reduce 
the imine in a single reaction. The selective reduction of iminium ions (but not carbonyl com- 
pounds) by sodium cyanoborohydride makes this possible. When NaCNBH 3 is added to a typical 
imine-formation reaction it reacts with the products but not with the starting carbonyl compound. 
Here is an example of an amine synthesis using reductive amination. 



u 

X 



NH 3 
NaCNBH 3 

pH 6 



NH 2 



PIT Me 

86% yield 



CH 2 =0 
NaCNBH 3 

pH 6 



1 


Ph^^Me 


81% yield 



In the first step, the ketone and ammonia are in equilibrium with their imine, which, at pH 6, is 
partly protonated as an iminium ion. The iminium ion is rapidly reduced by the cyanoborohydride 
to give the amine. Reactions like this, using ammonia in a reductive amination, are often carried out 
with ammonium chloride or acetate as convenient sources of ammonia. At pH 6, ammonia will be 
mostly protonated anyway. 



You will again meet the highly 
electrophilic iminium ions produced by 
reaction of formaldehyde with amines 
in Chapter 27, where we introduce you 
to the Mannich reaction. 





K 

PK Me 



u 

A. 



NH, 



PIT/ Me 
H 



^ CJT 

pH 6 Ph"C Me 

Of 

B 
H^I^CN 
H 

In the second step of the synthesis, amine plus formaldehyde gives an imine, present as its proto- 
nated iminium form, which gets reduced. Formaldehyde is so reactive that it reacts again with the 
secondary amine to give an iminium ion; again, this is reduced to the amine. 



Amines from imines: reductive amination 



355 



NH 2 


CH 2 =0 


i H N CN 

1 - 


H. .Me 

Ph^^Me 

secondary amine 


CH 2 =0 

*- 


H 
Me^/^CH 2 

©v 

► 


Me^ Me 

Ph^^Me 

tertiary amine 


Ph^^Me 

primary amine 


»- 

pH 6 


PIT Me 


»- 

pH 6 


Ph^^Me 



Living things make amino acids using imines 



The amino acid alanine can be made in moderate yield in 
the laboratory by reductive amination of pyruvic acid. 

Living things use a very similar reaction to manufacture 
amino acids from keto acids — but do it much more 
efficiently. The key step is the formation of an imine 
between pyruvic acid and the vitamin B6-derived amine 
pyridoxamine. 

Nature's synthesis of alanine: 
pyridoxamine 



A, 



Mr X0 2 H 

pyruvic acid 



NH 3 
NaCNBH 3 

9 

pH 6 



NH 2 



Me^ ^C0 2 H 50% yield 




Mr N C0 2 H 

pyruvic acid 

This Schiff base (biochemists call imines Schiff bases) is 
in equilibrium with an isomeric imine, which can be 
hydrolysed to pyridoxal and alanine. These reactions are, 
of course, all controlled by enzymes, and coupled to the 
degradation of unwanted amino acids (the latter process 



Me' X0 2 H 

alanine 

converts the pyridoxal back to pyridoxamine). Nature was 
doing reductive aminations a longtime before sodium 
cyanoborohydride was invented! We will come back to this 
in Chapter 50. 



An alternative method for reductive amination uses hydrogenation (hydrogen gas with a metal 
catalyst) to reduce the imine in the presence of the carbonyl compound. 

NH 3 

89% yield 



Ph„ 



,CH0 



high pressure required - 



70 °C 
H 2 , Ni - 
90 atm. 



N NH 2 

metal catalyst 



Lithium aluminium hydride reduces amides to amines 

We've talked about reduction of iminium ions formed from carbonyl compounds plus amines. 
Iminium ions can also be formed by reducing amides with lithium aluminium hydride. A tetrahedral 
intermediate is formed that collapses to the iminium ion. 




this metal could be aluminium or lithium: 
it's not important to the overall mechanism 

.metal 






N 
H 

tetrahedral 
intermediate 



iminium ion 




G 



Hydrogenation is a good way of 
reducing a number of different 
functional groups, but not (usually) 
carbonyl groups. In Chapter 24 we will 
look in more detail at reducing agents 
(and other types of reagent) that 
demonstrate selectivity for one 
functional group over another 
(chemoselectivity). 



'AIH, 



356 



14 ■ Nucleophilic substitution at C=0 with loss of carbonyl oxygen 



The iminium ion, is of course, more electrophilic than the starting amides (amide carbonyl 
groups are about the least electrophilic of any!), so it gets reduced to the secondary amine. This reac- 
tion can be used to make secondary amines, from primary amines and acyl chlorides. 






H,N 



/ r 






LiAIH 4